migsaa advanced ph.d. course

MIGSAA advanced Ph.D. courseDispersive Equations

UoE MATH11137Nonlinear Schrodinger Equations

Tadahiro Oh

Spring 2018

Contents:

• Lectures 1 - 3 by Tiffany Vlaar 3

• Lectures 4 - 5 by Albert Sola Vilalta 14

• Lectures 6 - 8 by Reuben Wheeler 25

• Lectures 9 - 10 by Felix James-Kahn 32

• Lectures 11 - 12 by Finlay Dupree McIntyre 39

• Lectures 13 - 15 by Andreia Chapouto 47

• Lectures 16 - 17 by Stefania Lisai 56

• Lectures 17 - 19 by William J. Trenberth 61

• Lectures 19 - 20 by Justin Forlano 68

LECTURES 1-3 DISPERSIVE EQUATIONS

TIFFANY VLAAR

1. Lecture 1

1.1. Introduction.

Some examples of nonlinear dispersive PDEs are

• Nonlinear Schrodinger equations (NLS)i∂tu+ ∆u = ±|u|p−1u,

u|t=0 = u0,(t, x) ∈ R× Rd, p > 1, (1.1)

where ∂t = ∂∂t , ∆ =

d∑j=1

∂2j =

d∑j=1

∂2

∂x2j, and u : (t, x) ∈ R× Rd → u(t, x) ∈ C.

• Nonlinear wave equations (NLW)

− ∂2t u+ ∆u = ±|u|p−1u, (t, x) ∈ R× Rd, (1.2)

where u is R-valued.

• Generalized Korteweg-de Vries equations (gKdV)

∂tu+ ∂3xu = ±∂x(up), (t, x) ∈ R× R, (1.3)

where is R-valued. If p = 2 this equation is known as the KdV equation. If p = 3it is known as the modified KdV (mKdV) equation.

Questions:

• Well-posedness?We say that (NLS) is locally well-posed (LWP) in Bs(Rd) (Banach space whoseelements have some degree of differentiability) if given any u0 ∈ Bs(Rd), thereexists an unique solution u to the (NLS) on [−T, T ] for some T = T (u0) > 0 withu|t=0 = u0 and with stability (as defined below). The same definition holds for(gKdV). The (NLW) however has a ∂2

t u term, and therefore requires initial data(u, ∂tu)|t=0 = (u0, u1) ∈ Bs ×Bs−1.

Stability means that it should be stable under some perturbation of the ini-tial data. In other words we require that the solution map

Φ : u0 ∈ Bs(Rd) 7→ u ∈ C(

[−T, T ];Bs(Rd))

is continuous for T = T (u0) > 0 small.

So local well-posedness is the existence of unique solutions for short timeswith stability. So an equation is locally well-posed if it satisfies the conditions of:existence, uniqueness, and continuity.

1

2 T. VLAAR

• If solutions exist for short times, then– Does the solution exist globally in time? If so, we say (NLS) is globally

well-posed.

– Otherwise, the solution for specific initial data may cease to exist at somefinite time ⇒ finite time blowup (formation of singularity)

• If GWP holds, then can we say anything about the behavior of global-in-timesolutions?e.g., scattering: a solution u(t) to (NLS) behaves asymptotically (as t→ ±∞) likea linear solution w : i∂tw + ∆w = 0, i.e., like a solution to the linear Schrodingerequation, but with different data⇒ ‖u(t)‖ → 0 as t→∞ in some appropriate norm.

• In some situations (i.e., if “s is too small”) we have ill-posedness results for theseequations.

If ill-posed, then what is the nature of the ill-posedness?– discontinuity of the solution map (often u0 = 0)– non-uniqueness– non-existence.

1.2. Background Materials.

In this section we discuss some general results which will be used throughout thecourse.

• Lebesgue spaces: Lp(X), X = Rd.

Lp(X) =

f on X : ‖f‖Lp(X)

def=

(ˆ

X|f(x)|pdx

)1/p

<∞

Note that ‖ · ‖Lp is a norm. Lp(X) is a normed vector space (p ≥ 1). And since itis also complete, it is a Banach space. If p = 2, it is a Hilbert space as we have theinner product

〈f, g〉L2 =

ˆ

Xfgdx.

• (X, dx), (Y, dy), measure spaces. Mixed norms

‖f‖LqyLrx =

∥∥∥∥∥∥‖

on X︷︸︸︷f(·, y)‖Lrx

∥∥∥∥∥∥Lqy

=

(ˆ

Y

(ˆ

X|f(x, y)|rdx

)q/rdy

)1/q

,

where by LqyLrx we denote the Lebesgue space Lq(Y ;Lr(X)).

LECTURES 1-3 DISPERSIVE EQUATIONS 3

• C(Rt;Lp(X)) =

f(t, x) : t 7→

function of x︷︸︸︷f(t) ∈ Lp(X) is continuous

as a map (R, | · |R) 7→(X, ‖ · ‖Lp(X)

).

Key Inequalities, Fourier Transforms and Schwartz spaces

(1) Holder’s inequality: If 1 ≤ p, q ≤ ∞ and 1p + 1

q = 1 (where q is called the Holder

conjugate of p), then

‖fg‖L1(X) ≤ ‖f‖Lp(X)‖g‖Lq(X) (1.4)

If p = q = 2 this is gives a form of the Cauchy-Schwarz inequality.

In general, if 1p + 1

q = 1r , then

‖fg‖Lr(X) ≤ ‖f‖Lp(X)‖g‖Lq(X) (1.5)

(2) Interpolation of Lp-spaces (log-convexity of Lp-norms)If 0 < p < q ≤ ∞ and θ ∈ [0, 1], then

‖f‖Lr ≤ ‖f‖θLp‖f‖1−θLq , where1

r=θ

p+

1− θq︸︷︷︸

p≤r≤q

(1.6)

(take logs to see convexity). If p =∞, then ‖f‖L∞(X) = ess supx∈X

|f(x)|.

(3) Minkowski’s integral inequality: If 1 ≤ p ≤ q ≤ ∞, then∥∥‖f‖Lp(X)

∥∥Lq(Y )

≤∥∥‖f‖Lq(Y )

∥∥Lp(X)

(1.7)

The usual Minkowski’s inequality follows from this by taking Lp(X) = l1(Z), thisgives: ‖ ∑

j∈Zfj‖Lq(Y ) ≤

∑j∈Z‖fj‖Lq(Y ).

(4) Young’s inequality: If 1r + 1 = 1

p + 1q , 1 ≤ p, q, r ≤ ∞, then

‖f ∗ g‖Lr ≤ ‖f‖Lp‖g‖Lq , (1.8)

where the convolution of f and g on Rd is defined as

f ∗ g(x) =

ˆ

Rdf(x− y)g(y)dy

=

ˆ

Rdf(y)g(x− y)dy.

(5) Riesz-Thorin interpolation theorem: If 1 ≤ pj , qj ≤ ∞, j = 0, 1, and T is a

linear operator defined on Lpj (Rd)→ Lqj (Rn) such that

‖Tf‖Lqj ≤ Aj‖f‖Lpj , j = 0, 1,

then

‖Tf‖Lq ≤ Aθ0A1−θ1 ‖f‖Lp ,

where θ ∈ [0, 1], 1p = θ

p0+ 1−θ

p1, 1q = θ

q0+ 1−θ

q1.

4 T. VLAAR

(6) Fourier transform on Rd. Consider f on Rd. The Fourier transform of f is

F(f)(ξ) = f(ξ) =

ˆ

Rdf(x)e−2πix·ξ dx, ξ ∈ Rd

Here x · ξ =d∑j=1

xjξj . This is well defined for f ∈ L1(Rd)

|f(ξ)| ≤ˆ

Rd|f(x)| dx = ‖f‖L1(Rd) <∞.

(7) The Schwartz class S(Rd) consists of all smooth functions whose derivatives arerapidly decreasing

S(Rd) = f ∈ C∞(Rd) : ‖f‖α,β = supx∈Rd|xα∂βf(x)| <∞,

where α, β are multi-indices, i.e., α = (α1, ..., αd), |α| =d∑j=1

αj and α ∈ (N ∪ 0)d.

For x ∈ Rd, xα = xα11 xα2

2 ...xαdd and ∂α = ∂α11 ∂α2

2 ...∂αdd , where ∂j = ∂∂xj

.

(8) Inverse Fourier transform

F−1(f)(x) = f(x)

def= f(−x) =

ˆ

Rdf(ξ)e2πix·ξ dξ.

Basic properties for f ∈ S(Rd)• ‖f‖L2 = ‖f‖L2 = ‖f‖L2 , Plancherel’s identity• F : L2 → L2, bijection• (f)∨ = f = (f)∧

• Parseval’s identityˆ

f(x)g(x) dx =

ˆ

f(ξ)g(ξ) dξ

〈f, g〉L2x

= 〈f , g〉L2ξ.

(9) Hausdorff-Young’s inequality: If 1p + 1

p′ = 1, p ≥ 2, then

‖f‖Lp ≤ ‖f‖Lp′This follows from Plancherel’s identity (see (8)) and the Riesz-Thorin interpolationtheorem (see (5)).

2. Lecture 2

Reading Material:

• Folland, Real Analysis (Chapter 6,8,9)• Grafakos, Classical Fourier Analysis (Chapter 1,2,5), Modern Fourier Analysis

(Chapter 6)• Tao, Nonlinear Dispersive PDEs (Appendix)• Cazenave, Semilinear Schrodinger Equations• Linares-Ponce, Introduction to Nonlinear Dispersive PDEs.


Fourier transform

(∂αf)∧(ξ) = (2πiξ)αf(ξ)

(∂αf)∨(x) = (−2πix)αf(x)

f ∗ g(ξ) = f(ξ)g(ξ), f ∗ g = fg

Dilation property. Let fε(x) = 1εdf(xε

), then fε(ξ) = f(εξ). Let gε(x) = g

(xε

), then

gε(ξ) = εdg(εξ).

Lemma 2.1 (Riemann-Lebesgue Lemma). Let f ∈ L1(Rd). Then f(ξ) → 0 as |ξ| → ∞.It also holds that

F(L1(Rd)) ⊂ C0(Rd),where C0 is the space of continuous functions that decay to zero at infinity. Cc is the spaceof continuous functions with compact support.

Denote the space of tempered distributions by S ′(Rd). T ∈ S ′(Rd) : S → C is linearand continuous. From the topology defined on the Schwartz space and its dual, one obtainsthat Tk → T in S ′, if Tk(f)→ T (f), ∀f ∈ S. Weak-* topology.

One can define a Fourier Transform on S ′:f ∈ S ′, g ∈ S ⇒ f(g)“ = 〈f , g〉 ”

def= f(g)

Sobolev space: Hs(Rd) = completion of S with respect to

‖f‖Hs =

(ˆ

(1 + |ξ|2)s|f(ξ)|2 dξ)1/2

, s ∈ R

= ‖〈ξ〉sf(ξ)‖L2ξ(Rd),

where 〈ξ〉 is used to denote (1 + |ξ|2)1/2.

Remarks

• From this one obtains from Plancherel H0 = L2.

• H1 = L2 ∩ ∂f ∈ L2 ⇐ recall |ξ|2|f(ξ)|2 = |∇f(ξ)|2 (up to a constant).

‖f‖2H1 = ‖f‖2L2 + ‖∇f‖2L2

• General, for s > 0 : Hs = f ∈ L2 and derivatives up to order s are also in L2.For s < 0 : 〈∇〉sf lies in L2.

Theorem 2.2 (Sobolev Embedding Theorem). If s > d2 , then Hs(Rd) ⊂ L∞(Rd)

(and continuously embedded, ‖f‖L∞ ≤ C‖f‖Hs , ∀f ∈ Hs and some constant C).

6 T. VLAAR

Proof.ˆ

f(ξ)e2πix·ξ dξ = ‖f‖L∞x ≤ ‖f‖L1ξ

=

ˆ

〈ξ〉−s〈ξ〉s|f(ξ)| dξ

Cauchy-Schwarz≤

(ˆ

1

〈ξ〉2s dξ)1/2

‖f‖Hs

≤ C‖f‖Hs if s >d

2.

Notations

• A . B if A ≤ CB for some C > 0• A ∼ B if A . B and B . A• A B if A ≤ εB for some small ε > 0.

Other Sobolev spaces

(1) Lps = Bessel potential space,

‖f‖Lps = ‖F−1(〈ξ〉sf(ξ)

)‖Lpx .

The Sobolev embedding holds for ps > d. If p = 2: L2s = Hs.

(2) W s,p, 0 < s < 1

‖f‖W s,p = ‖f‖Lp +

(ˆ ˆ |f(x)− f(y)|p

|x− y|d+spdxdy

)1/p

See Stein’s book: singular integral, Λps.

(3) We define the inhomogeneous Sobolev space Hs(Rd) as the completion of S(Rd)with respect to the following norm

‖f‖Hs :=

(ˆ

Rd(1 + |ξ|2)s|f(ξ)|2 dξ

)1/2

Similarly, the homogeneous Sobolev space Hs(Rd) is defined as the completion ofS(Rd) with respect to the following norm

‖f‖Hs :=

(ˆ

Rd|ξ|2s|f(ξ)|2 dξ

)1/2

Example: H1 = ∇f ∈ L2 (no need to have f ∈ L2).

Denote by Lps the Riesz potential space.

Remark 2.3. For s ≥ 0 : Hs ⊂ Hs.

Remark 2.4. Hs “norm” is in fact a seminorm. Hs is therefore defined up topolynomial (s ≥ 0).


Sobolev Inequality: If 0 ≤ sd = 1

p − 1q , 1 < p ≤ q <∞, then

‖f‖Lq ≤ C‖f‖W s,p ,

i.e., one “controls higher order integrability by the Lp-norm of the s-derivative of f”.

Convention: We use W s,p for Lps (these spaces are equivalent for p ∈ Z), i.e., forus

‖f‖W s,p = ‖F−1(〈ξ〉sf(ξ)‖Lpx

Interpolation: If s = θs1 + (1− θ)s2, θ ∈ (0, 1), then

‖f‖Hs ≤ ‖f‖θHs1‖f‖1−θHs2

Algebra property of Hs: If s > d2 , H

s ⊂ F(L1) ⊂ C0 (by Riemann-Lebesgue), then

f, g ∈ Hs ⇒ fg ∈ Hs (pointwise product ⇐ ‖fg‖Hs . ‖f‖Hs‖g‖Hs).

Proof. 〈ξ〉s . 〈ξ−ξ1〉s+〈ξ1〉s(

by triangle inequality and (a+ b)θ ≤aθ + bθ, θ ∈ (0, 1]

caθ + cbθ, θ > 1

)

‖fg‖Hs = ‖〈ξ〉sfg(ξ)‖L2ξ, where fg(ξ) = f ∗ g(ξ) =

ˆ

f(ξ − ξ1)g(ξ1)dξ1

.∥∥∥∥ˆ (〈ξ − ξ1〉s|f(ξ − ξ1)|

)|g(ξ1)|dξ1

∥∥∥∥L2ξ

+

∥∥∥∥ˆ

|f(ξ − ξ1)| (〈ξ1〉s|g(ξ1)|) dξ1

∥∥∥∥L2ξ

=∥∥∥(〈ξ〉s|f |

)∗ |g|

∥∥∥L2ξ

+∥∥∥|f | ∗ (〈ξ〉s|g|)

∥∥∥L2ξ

Young’s inequality≤

∥∥∥〈ξ〉sf∥∥∥L2ξ

‖g‖L1ξ

+∥∥∥f∥∥∥L1ξ

‖〈ξ〉sg‖L2ξ

≤ ‖f‖Hs‖g‖L1ξ

+ ‖f‖L1ξ‖g‖Hs

. ‖f‖Hs‖g‖Hs (from the Sobolev Embedding theorem proof).

This algebra property of the Sobolev spaces will be used to prove the local well-posednessof the nonlinear Schrodinger equation in (Hs, Hs−1).

3. Lecture 3

3.1. Local Well-Posedness of NLS in Hs(Rd), s > d/2.

First consider the linear Schrodinger equationi∂tu+ ∆u = 0,

u|t=0 = u0.(3.1)

8 T. VLAAR

Taking the Fourier Transform (F.T.) in x then givesi∂tu(ξ)− |ξ|2u(ξ) = 0,

u(ξ)|t=0 = u0(ξ),(3.2)

because ∆ = ∂21 + ...+∂2

dF.T.−→ ∆(ξ) = −4π2(ξ2

1 + ...+ξ2d) = −4π2|ξ|2, where we have omitted

the constant factor 4π2 for simplicity. For fixed ξ, u(t, ξ) = e−it|ξ|2u0(ξ). This gives

u(t, x) = F−1(e−it|ξ|

2u0(ξ)

)(x)

=:(eit∆u0

)(x),

where S(t) = eit∆ is the linear Schrodinger operator.

Now consider the inhomogeneous linear Schrodinger equationi∂tu+ ∆u = F (t, x),

u|t=0 = u0,(3.3)

where F is given (and “nice”). Taking the F.T. in x then gives

∂tu(ξ) + i|ξ|2u(ξ) = −iF (t, ξ), ∀ξ ∈ Rd

⇒ ∂t

(eit|ξ|

2u(ξ)

)= −ieit|ξ|2F (t, ξ).

Integrate this from 0 to t to obtain

eit|ξ|2u(t, ξ)− u0(ξ) = −i

ˆ t

0eit′|ξ|2F (t′, ξ) dt′

⇒ u(t, ξ) = e−it|ξ|2u0(ξ)− i

ˆ t

0e−i(t−t

′)|ξ|2F (t′, ξ) dt′

Take F−1

⇒ u(t) = S(t)u0 − iˆ t

0S(t− t′)F (t′) dt′.

Back to NLS with nonlinearity N (u, u) = up1 up2 , where p1 + p2 = p and pj ∈ N ∪ 0. Wesay that u is a solution to (NLS) with nonlinearity N if u satisfies the following Duhamelformulation:

u(t) = S(t)u0 − iˆ t

0S(t− t′)N (u, u)(t′) dt′

=: Γu0(u)(t),

Goal: Given u0 ∈ Hs(Rd), show that there exists an unique u such that u = Γu0(u) on[−T, T ], T = T (u0) > 0, i.e., u is a fixed point of Γu0 .

Basic properties of S(t)

(1) S(t) is unitary on Hs(Rd), s ∈ R.

Proof.

‖S(t)f‖Hs =

(ˆ

〈ξ〉2s∣∣∣e−it|ξ|2 f(ξ)

∣∣∣2dξ

)1/2

= ‖f‖Hs

so the operator S(t) preserves length.


(2) S(t)f ∈ C(Rt;Hsx(Rd)), f ∈ Hs,

S(·↓)f : t 7→ S(t)f ∈ Hs continuous

Hs-valued function of t

Proof. Fix t ∈ R. Then using the semigroup property S(t1 + t2) = S(t1)S(t2)

‖S(t+ h)f − S(t)f‖Hs = ‖S(t)(S(h)− I )f‖Hs

= ‖(S(h)− I )f‖Hs ,

where the last step follows from the unitarity property (1).

Separate on the Fourier side

• Take |ξ| > N such that∥∥∥F−1

(I |ξ|>Nf(ξ)

)∥∥∥Hs

< ε4 .

• For |ξ| ≤ N , use the mean value theorem

|(S(h)− 1)∧(ξ)| = |e−ih|ξ|2 − 1| ≤ N2|h|.

Notation: CTHs = C([−T, T ];Hs(Rd))

and ‖u‖CTHs = ‖‖u(t)|Hs‖L∞T , where L∞T = L∞([−T, T ]).

Fix u0 ∈ Hs(Rd), s > d2 , then

‖Γu0(u)‖CTHs ≤ ‖S(t)u0‖CTHs +

∥∥∥∥ˆ t

0S(t− t′)N (u, u)(t′) dt′

∥∥∥∥CTHs

≤ ‖u0‖Hs +

ˆ T

0

independent of t︷︸︸︷‖N (u, u)‖CTHs dt (from unitarity of S(t) & Minkowski’s integral ineq.)

≤ ‖u0‖Hs + T‖up1 up2‖CTHs

≤ ‖u0‖Hs + C‖u‖p1CTHs‖u‖p2CTHs︸︷︷︸=‖u‖pCTHs

we want this to be ≤ 2‖u0‖Hs =: R.

For u ∈ BR = closed ball of radius R in CTHs

‖Γu0(u)‖CTHs ≤ ‖u0‖Hs + CTRp

≤ R

2+ (2CTRp−1) · R

2≤ R,

where 2CTRp−1 ≤ 1 by choosing T ≤ (2CRp−1)−1. Note: T → 0 as initial data gets larger.

It follows that Γu0 : BR → BR. So the map Γu0 maps the ball BR onto itself.

10 T. VLAAR

Now show the contraction. For u, v ∈ BR

‖Γu0(u)− Γu0(v)‖CTHs ≤ˆ T

0‖N (u)−N (v)‖CTHs dt.

Note that

up11 up2 − vp1vp2 telescoping sum

= (u− v)up1−1up2 + v(u− v)up1−2up2 + ...+ vp1 vp2−1(u− v).

Plugging this into the above equation gives

‖Γu0(u)− Γu0(v)‖CTHs ≤ CT

p−1∑

j=0

‖u‖p−1−jCTHs ‖v‖jCTHs

‖u− v‖CTHs .

From Young’s inequality: 1p + 1

q = 1 ⇒ ab ≤ ap

p + bq

q . For j = 1 with p−2p−1 + 1

p−1 = 1, one

gets ‖u‖p−2CTHs‖v‖CTHs ≤ C

(‖u‖(p−2) p−1

p−2

CTHs + ‖v‖p−1CTHs

). By performing Young’s inequality

on each term one then gets

‖Γu0(u)− Γu0(v)‖CTHs ≤ CT(‖u‖p−1

CTHs + ‖v‖p−1CTHs

)‖u− v‖CTHs

≤ CTRp−1‖u− v‖CTHs

≤ 1

2‖u− v‖CTHs ,

where the last inequality follows by choosing T ≤ 1

2CRp−1.

Summary: ∀u, v ∈ BR

‖Γu0(u)‖CTHs ≤ R

and ‖Γu0(u)− Γu0(v)‖CTHs ≤ 1

2‖u− v‖CTHs ,

by choosing T ≤ min(

12CRp−1 ,

1

2CRp−1

).

Recall that R = 2‖u0‖Hs , so T = T (‖u0‖Hs) > 0.

Theorem 3.1 (Banach fixed point theorem (Contraction mapping principle)). A contrac-tion (namely T : BR → BR and ‖T (u) − T (v)‖ ≤ θ‖u − v‖, where θ < 1) on a closed ballin a complete metric space has a unique fixed point.

So the Banach fixed point theorem tells us that there exists an unique solution u ∈ BR

such that u = Γu0(u). Namely, u is a solution to NLS.

Remarks

(1) u ∈ CTHs (⇐ requirement for LWP)

u(t) = S(t)u0︸︷︷︸∈CTHs

− iˆ t

0S(t− t′)N (u, u)(t′) dt′

︸︷︷︸G(t)

.


Want to show time continuity. Use telescoping sum to obtain

G(t+ h)−G(t) =

ˆ t+h

0S(t+ h− t′)N (u, u)(t′) dt′ −

ˆ t

0S(t− t′)N (u, u)(t′) dt′

=

ˆ t+h

tS(t+ h− t′)N (u, u)(t′) dt′

︸︷︷︸small by shortness of [t,t+h]

−ˆ t

0

(S(t+ h− t′)− S(t− t′)

)︸︷︷︸

=S(t−t′)(S(h)−1) by semigroup property

N(u, u)(t′) dt′.

(2) At this point, u is unique only in BR ⊂ CTHs. In some cases it is possible to proveuniqueness of u in CTH

s (possibly by shrinking T = T (R) a bit).Method 1: Gronwall’s Inequality. Method 2: Bootstrap argument.

LECTURES 4 AND 5

ALBERT SOLA VILALTA

1. Conservation laws, global well-posedness and persistence of regularity

We first study three conservation laws for (NLS), namely conservation of mass, conserva-tion of energy (Hamiltonian) and conservation of momentum. We use them to show globalwell-posedness in H1(R) of the 1 dimensional defocusing (NLS) with odd power p. Wethen study persistence of regularity and show global well-posedness of the 2 dimensionaldefocusing cubic (NLS) in H2(R2).

Before continuing, we recall the Nonlinear Schrodinger equation (NLS) we are looking at

i∂tu+ ∆u = ±|u|p−1u

u(0, x) = u0(x)(1.1)

where p > 1, u0 ∈ Hs(Rd) is the initial data and u : R × Rd → C is the unknown. Thedefocusing or repulsive (resp. focusing or attractive) case corresponds to the + (resp. −)sign in (1.1).

1.1. Conservation laws. Let u ∈ H1(Rd;C). We define its mass M(u) as

M(u) :=

ˆ

Rd|u(x)|2dx; (1.2)

its Hamiltonian H(u), also called energy, as

H(u) :=1

2

ˆ

Rd|∇u|2dx± 1

p+ 1

ˆ

Rd|u|p+1dx; (1.3)

and its momentum P (u) as

P (u) := Im

ˆ

Rdu∇udx = −i

ˆ

Rdu∇u; (1.4)

where we used integration by parts in the last equality.

Proposition 1.1 (Conservation of mass, energy and momentum). Let s > d2 , u0 ∈ Hs(Rd)

and u ∈ CTHs be a solution to (NLS). The mass, energy and momentum of u are conserved,i.e., for all t ∈ [0, T ], we have

M(u(t)) = M(u0),

H(u(t)) = H(u0)

and

P (u(t)) = P (u0)

We will show only the conservation of mass, leaving the rest as an exercise for the reader.1

2 ALBERT SOLA VILALTA

Proof. Consider first the case u0 ∈ H∞(Rd). Let u ∈ CTH∞ be the unique solution to

(NLS) with initial data u0. We have,

∂t

ˆ

Rd|u|2 = 2 Re

ˆ

Rd(∂tu)u = 2 Re i

ˆ

Rd(∆u)u∓ 2 Re i

ˆ

Rd|u|p+1

= −2 Re i

ˆ

Rd|∇u|2 + 0 = 0

For rough solutions, we use well-posedness theory (continuous dependence) to prove theconservation of mass. Consider now u0 ∈ Hs(Rd). Let u0,n∞n=1 ⊂ H∞(Rd) be such that

u0,n → u0 in Hs(Rd) (1.5)

Let un ∈ CTHs be the solution to (NLS) with initial data u0,n, i.e., un satisfies the Duhamelformulation

un(t) = S(t)u0,n ∓ iˆ t

0S(t− t′)|un(t′)|p−1un(t′)dt′

and u ∈ CTHs be the solution to (NLS) with u(0, ·) = u0, i.e., u satisfies the Duhamel

formulation

u(t) = S(t)u0 ∓ iˆ t

0S(t− t′)|u(t′)|p−1u(t′)dt′

For sufficiently large n, we have

||u0,n||Hs ≤ ||u0||Hs + 1 =:R

2

The local well-posedness obtained in Lecture 3 gives the existence of T = T (R) > 0 suchthat un and u exist on [−T, T ]. Recall that T depends only on the Hs-norm of the initialdata, so (1.5) ensures that the same T is valid for u and all un with n ≥ N , where N is asufficiently large integer. Now, using Duhamel’s formulation

||un − u||CTHs ≤ ||u0,n − u0||Hs +

ˆ T

0

∣∣∣∣∣∣|un|p−1un − |u|p−1u

∣∣∣∣∣∣CTHs

≤ ||u0,n − u0||Hs + CTRp−1||un − u||CTHs

where we did analogous computations to those in the proof of LWP, see Lecture 3. Bychoosing T = T (R) small, i.e., such that CTRp−1 ≤ 1

2 , we get

||un − u||CTHs . ||u0,n − uo||Hs (1.6)

Finally,

M(u(t))12 = ||u(t)||L2 ≤ ||u(t)− un(t)||L2 + ||un(t)||L2

≤ ||u− un||CTHS + ||u0,n||L2

≤ C||u0 − u0,n||Hs + ||u0,n||L2 → 0 + ||u0||L2 = M(u0)12 as n→∞

(1.7)

where we used conservation of mass for smooth functions and (1.6). Analogously, by chang-ing the roles of 0 and t, one obtains M(u0) ≤ M(u(t)). Therefore, we conclude that themass is conserved.

In the next Theorem, we use local well-posedness and the conservation of mass andenergy to show global well-posedness of the 1 dimensional defocusing (NLS).

LECTURES 4 AND 5 3

Theorem 1.2 (Global Well-Posedness in H1(R)). Let p ∈ 2N+1. Then, the 1 dimensionaldefocusing (NLS)

i∂tu+ ∆u = |u|p−1u

u(0, x) = u0(x)(1.8)

with initial data u0 ∈ H1(R) is globally well-posed in H1(R). Here u : R× R→ C denotesthe unknown.

Proof. We first observe that

||u(t)||2H1 = ||u(t)||2L2 + ||∇u(t)||2L2

≤ c(M(u(t)) +H(u(t))) = c(M(u0) +H(u0)) =: C(u0)2 <∞(1.9)

by the conservation of mass and energy. Notice that the quantity H(u0) is finite by Sobolevembedding. Set R := 2C(u0), which is greater or equal than 2||u(t)||H1 . Apply the LWPargument with this R. We get the existence of a solution up to time T = T (R).

We iterate this process. Consider now initial data uT := u(T, ·). From (1.9), it followsthat ||uT ||H1 ≤ C(u0), so if we choose R = 2C(u0) again and apply the LWP argument,we get the existence of a solution up to time 2T . Proceed in this fashion to get a solutiondefined at all times. We stress that T depends only on R, which by (1.9) can be taken tobe the same at every iteration of the LWP argument.

1.2. Persistence of regularity. A natural question that arises after Theorem 1.2 is whathappens if we increase the regularity of initial data. More precisely, let s > 1 and assumeu0 ∈ Hs(R). Since, in particular, u0 ∈ H1(R), Theorem 1.2 gives GWP in H1(R), i.e.,u ∈ C(R;H1(R)).

Question: Can we also conclude u ∈ C(R;Hs(R))?The answer to this question is yes, as we shall see in this section. We begin by recalling

a product estimate, which follows from the paraproduct formula. Let s ≥ 0 and assumef, g ∈ Hs(Rd). Then,

||fg||Hs(Rd) . ||f ||L∞(Rd)||g||Hs(Rd) + ||f ||Hs(Rd)||g||L∞(Rd) (1.10)

Proposition 1.3 (Persistence of regularity). Let p ∈ 2N + 1. Then, the 1 dimensionaldefocusing (NLS)

i∂tu+ ∆u = |u|p−1u

u(0, x) = u0(x)(1.11)

with initial data u0 ∈ Hs(R) is globally well-posed in Hs(R).

Proof. We have

|||u|p−1u||Hs(R) . ||u||p−1L∞(R)||u||Hs(R) . ||u||p−1

H1(R)||u||Hs(R) . Rp−1||u||Hs(R)

where we used (1.10) successively in the first inequality and Sobolev embedding in thesecond. Here R is the same quantity as in the proof of Theorem 1.2. Duhamel’s formulationand the previous estimate give,

||u(t)||Hs(R) ≤ ||u0||Hs(R)+

ˆ t

0|||u(t′)|p−1u(t′)||Hs(R)dt

′ ≤ ||u0||Hs(R)+cRp−1

ˆ t

0||u(t′)||Hs(R)dt

′

From Gronwall’s inequality, we conclude

||u(t)||Hs(R) ≤ ecRp−1t||u0||Hs(R) ≤ ∞. (1.12)


Therefore, u(t) ∈ Hs(R) for all t ∈ [0,∞). From LWP, we deduce u ∈ C(R;Hs(R)).

Remark 1.4. Estimate (1.12) provides an exponential upper bound on the growth of asolution. It is of importance to improve this bound. In the mid 90’s, Bourgain and Staffilaniobtained the polynomial bound

||u(t)||Hs(R) ≤ C(u0)tα(s−1)

Question: Can we obtain a subpolynomial upper bound?Question: Can we construct a solution s.t. the Hs-norm actually grows?

In what follows, we study global well-posedness of the 2 dimensional defocusing cubic(NLS) in H2(R2). Recall that, since we are in two dimensions, we have LWP in Hs(R2) fors > 1 = d

2 . We begin by showing the Brezis-Gallouet inequality, which will be used in theproof of GWP.

Proposition 1.5 (Brezis-Gallouet inequality, ’80). Let s > d2 . Then,

||f ||L∞(Rd) ≤ Cs||f ||H d2 (Rd)

[log

(2 +||f ||Hs(Rd)

||f ||Hd2 (Rd)

)] 12

(1.13)

Remark 1.6. Brezis-Gallouet inequality can be seen as a refinement of Sobolev’s EmbeddingTheorem in the limiting case s = d

2 . We recall Sobolev’s Embedding Theorem, which states

that for s > d2 ,

||f ||L∞(Rd) . ||f ||Hs(Rd)

Proof. We first note that it suffices to show

||g||L∞(Rd) ≤ Cs(

log(2 + ||g||Hs(Rd))) 1

2 (1.14)

for g ∈ Hs(Rd) such that ||g||Hd2 (Rd)

= 1. Indeed, for general f ∈ Hs(Rd), we have

||f ||L∞(Rd) = ||f ||Hd2 (Rd)

∣∣∣∣∣∣∣∣

f

||f ||Hd2 (Rd)

∣∣∣∣∣∣∣∣L∞(Rd)

≤ Cs||f ||Hd2 (Rd)

[log

(2 +||f ||Hs(Rd)

||f ||Hd2 (Rd)

)] 12

.

Therefore, consider g ∈ Hs(Rd) such that ||g||Hd2 (Rd)

= 1. We have,

||g||L∞(Rd) ≤ ||g||L1(Rd) =

ˆ

|ξ|≤R|g(ξ)|dξ +

ˆ

|ξ|>R|g(ξ)|dξ

=

ˆ

|ξ|≤R〈ξ〉 d2 |g(ξ)| 1

〈ξ〉 d2dξ +

ˆ

|ξ|>R〈ξ〉s|g(ξ)| 1

〈ξ〉sdξ

≤ ||g||Hd2 (Rd)

(ˆ

|ξ|≤R

dξ

〈ξ〉d) 1

2

+ ||g||Hs(Rd)

(ˆ

|ξ|>R

dξ

〈ξ〉2s) 1

2

∼(

log(2 +R)) 1

2 +Rd2−s||g||Hs(Rd)

Set R = ||g||θHs(Rd)

with θ := 1s− d

2

> 0. We have shown (1.14), which concludes the

proof.

LECTURES 4 AND 5 5

Theorem 1.7 (Global Well-Posedness in H2(R2)). The 2- dimensional defocusing cubic(NLS)

i∂tu+ ∆u = |u|2uu(0, x) = u0(x)

(1.15)

with initial data u0 ∈ H2(R2) is globally well-posed in H2(R2). Here u : R × R2 → Cdenotes the unknown.

Proof. Using Duhamel’s formulation, Minkowski’s integral inequality, the product estimate(1.10) and Brezis-Gallouet inequality with s = 2 > d

2 = 1, we get

||u(t)||H2(R2) ≤ ||S(t)u0||H2(R2) +

∣∣∣∣∣∣∣∣ˆ t

0S(t− t′)|u|2u(t′)dt′

∣∣∣∣∣∣∣∣H2(R2)

≤ ||u0||H2(R2) +

ˆ t

0|||u(t′)|2u(t′)||H2(R2)dt

′

. ||u0||H2(R2) +

ˆ t

0||u(t′)||H2(R2)||u(t′)||2L∞(R2)dt

′

≤ ||u0||H2(R2) + c

ˆ t

0||u(t′)||H2(R2)(1 + log(1 + ||u(t′)||H2(R2)))dt

′ =: F (t)

where the constant c depends on ||u(t)||H1(R2) appearing in the Brezis-Gallouet inequality.From the estimate

F ′(t) = ||u(t)||H2(R2)(1 + log(1 + ||u(t)||H2(R2))) ≤ F (t)(1 + log(1 + F (t)))

we deduce that

d

dt[log(1 + log(1 + F (t)))] =

1

1 + log(1 + F (t))

1

1 + F (t)F ′(t) ≤ 1

which gives

||u(t)||H2(R2) ≤ F (t) ≤ α(u0)eeβt

for some constants α, β. This concludes the proof.

2. Scaling heuristics

In this section we study the symmetries of (NLS), with emphasis on the scaling symmetrybecause it characterises whether the Cauchy Problem for (NLS) is subcritical, critical orsupercritical (w.r.t. scaling). We will make these concepts clear below.

Consider the (NLS) with power nonlinearity,

i∂tu+ ∆u = ±|u|p−1u on R× Rd

u(0, x) = u0(x)(2.1)

where u0 ∈ Hs(Rd) for some s > 0 denotes the initial data. Consider the scaled initial data

uλ0(x) :=1

λau0

(xλ

)

and the scaled solution

uλ(t, x) :=1

λau( t

λ2,x

λ

)

where a ∈ R satisfies a+ 2 = ap, i.e., a = 2p−1 .


A simple computation shows that if u is a solution to (2.1) with initial data v0, then uλ

is also a solution to (2.1) with initial data vλ0 . Indeed,

i∂tuλ + ∆uλ =

1

λ2+a

[i∂tu

( t

λ2,x

λ

)+ ∆u

( t

λ2,x

λ

)]=

1

λ2+a

∣∣∣u( t

λ2,x

λ

)∣∣∣p−1

u( t

λ2,x

λ

)

= |uλ|p−1uλ

This scaling symmetry induces the so-called scaling-critical Sobolev index sc = scrit,defined to be the exponent such that

||fλ||Hsc (Rd) = ||f ||Hsc (Rd)

holds. Notice that here f : Rd → R and fλ : Rd → R is defined by fλ(x) := 1λa f(xλ) with

a = 2p−1 , i.e., we only look at the space variable. We determine sc now.

A change of variables gives fλ(ξ) = λd−af(λξ). We have,

||fλ||2Hs(Rd)

=

ˆ

Rd|ξ|2sλ2(d−a)|f(λξ)|2dξ =

ˆ

Rd

∣∣∣ηλ

∣∣∣2sλ2(d−a)λ−d|f(η)|2dη

= λd−2s−2a||f ||2Hs(Rd)

(2.2)

Therefore, sc satisfies the relation

0 = d− 2sc − 2a = d− 2sc − 22

p− 1

which gives

sc =d

2− 2

p− 1. (2.3)

Definition 2.1 (Criticality of (NLS) w.r.t. scaling). Given u0 ∈ Hs(Rd), the Cauchyproblem (2.1) w.r.t. scaling is

• subcritical if s > sc• critical if s = sc• supercritical if s < sc.

In the subcritical case, good behaviour is expected. By this, we mean LWP, possibilityto extend it to GWP, persistence of regularity, etc. In the critical case, there is a delicatebalance between linear dispersion and nonlinear concentration and therefore the argumentsmight be more involved and at the end one might not be able to recover all the propertiesfrom the subcritical case. Finally, in the supercritical case, ill-posedness is expected. Inwhat follows, we describe this heuristics further.

Let λ >> 1 1. From (2.2), we have

||uλ0 ||Hs(Rd) = λsc−s||u0||Hs(Rd)

and the time-scaling t 7−→ tλ2

gives that if u exists on the time interval [0, T ], then uλ exists

on the time interval [0, T λ] := [0, λ2T ]. The combination of these two conditions gives theheuristics.

In the subcritical case, since sc−s < 0, we have ||uλ0 ||Hs(Rd) << ||u0||Hs(Rd) and T λ >> T .

This is, if the initial data gets smaller, the time up to which solutions are defined increases.This is coherent with the proof of LWP, where the time up to which solutions are definedis essentially inversely proportional to the Hs(Rd)-norm of the initial data. In short, in

1An analogous discussion can be done if λ << 1. λ ∼ 1 would give essentially nothing.

LECTURES 4 AND 5 7

the subcritical case the solution behaves well w.r.t. scaling and therefore we expect goodbehaviour.

In the supercritical case, since sc − s > 0, we have ||uλ0 ||Hs(Rd) >> ||u0||Hs(Rd) and

T λ >> T . This is, if the initial data gets bigger, the time up to which solutions are definedincreases. Note that this is the opposite of what the proof of LWP hints, and therefore weexpect something to break for sufficiently big λ. Thus, we do not expect good behaviourin the supercritical case.

In the critical case, we have ||uλ0 ||Hs(Rd) = ||u0||Hs(Rd) and T λ >> T . This is, scaling

does not increase initial data but does increase the time up to which solutions are defined.In this case, we have to be more careful because the increase of existence time could beincoherent, as in the supercritical case, or could be justified by having GWP. Therefore, inthe critical case we need more information than the Hs-norm of the initial data.

We conclude the section with other symmetries of (NLS).

• Time translation: t 7→ t+ t0• Spatial translation: x 7→ x+ x0

• Rotation: u 7→ eiθu for all θ ∈ R• Galilean symmetry: u(t, x) 7→ ei

v2·xe−i

|v|24tu(t, x+ vt)

• Time reversal: u(t, x) 7→ u(−t, x)

Note that the Galilean symmetry induces another critical regularity s∞c = 0.

Remark 2.2. If we prove LWP by a fixed point argument (Contraction Mapping Principle)for |u|p−1u, p ∈ 2N + 1 (algebraic), then the solution map

Φ : u0 ∈ Hs(Rd) 7−→ u ∈ CTHs

is analytic. Therefore, if we know that the solution map is not smooth, it means that wecan not prove LWP by a fixed point argument (but does not say that it is ill-posed, anotherargument to prove LWP might still be possible).

We will discuss more about ill-posedness later in the course.

3. Strichartz estimate

We begin by studying how the linear Schrodinger operator S acts on some waves. Con-sider the Linear Schrodinger equation

i∂tu+ ∆u = 0 in R× Rd

u(0, x) = u0(x)(3.1)

Applying the space Fourier transform to (3.1) gives the solution

u(t, x) = F−1(e−4π2it|·|2 u0(·))(x) =: S(t)u0(x)

Writing down S(t)u0 explicitly

S(t)u0(x) =

ˆ

Rde2πix·ξe−4π2it|ξ|2 u0(ξ)dξ =

ˆ

Rde2πiξ·(x−2πξt)u0(ξ)dξ (3.2)

we see that if u0 is localised around ξ = ξ0 ∈ Rd, then S(t)u(x) has phase velocity ∼ 2πξ0.Consider u0(x) := e2πix·ξ0 , i.e., a plane wave at frequency ξ0. We note that it is not an

L1(Rd) nor L2(Rd) function and therefore we see its Fourier transform as a distribution. Infact,

F(e2πix·ξ0)(ξ) = δξ0(ξ),


the Dirac delta distribution at ξ0. Let φ ∈ S(Rd) be a test function. We have,

〈S(t)u0(x), φ〉 = 〈e−4π2it|ξ|2δξ0 ,F−1φ〉 = 〈δξ0 , e−4π2it|ξ|2F−1φ〉 =

ˆ

Rde−4π2it|ξ0|2e2πix·ξ0φ(x)dx

which means that S(t)u0(x) is the function

S(t)u(x) = e−4π2it|ξ0|2e2πix·ξ0 = e2πiξ0(x−2πξ0t)

Note that in this case, i.e., when supp u0(ξ) = ξ0, the phase velocity is effectively 2πξ0,justifying the previous heuristics. We look now at the following spatially localised wave

v0(x) = e−|x|24σ2 e2πix·ξ0

with Fourier transform given by

v0(ξ) = (4πσ2)d2 e−4π2σ2|ξ−ξ0|2 . (3.3)

In order to compute the Fourier transform of v0 we can use the fact that the Fouriertransform of a Gaussian is a Gaussian. We state it as a separate Lemma.

Lemma 3.1 (Fourier transform of a Gaussian). Let

g(x) := e−π|x|2.

Then,

g(ξ) = e−π|ξ|2

Proof.

g(ξ) =

ˆ

Rde−π|x|

2e−2πix·ξdx =

d∏

j=1

ˆ

Re−πx

2j−2πixjξjdxj =

d∏

j=1

e−πξ2j

ˆ

Re−π(xj+iξj)

2dxj

=d∏

j=1

e−πξ2j

ˆ

Re−πy

2dy = e−π|ξ|

2

Using Lemma 3.1, we can compute the Fourier transform of v0. Let

ν0(x) := e−|x|24σ2 ,

which satisfies v0(ξ) = ν0(ξ − ξ0) and

ν0(ξ) =

ˆ

Rde−π∣∣ x

(4πσ2)1/2

∣∣e−2πix·ξdx = (4πσ2)

d2

ˆ

Rde−π|y|

2e−2πiy·(4πσ2)1/2ξdy

= (4πσ2)d2 e−4π2σ2|ξ|2

where we used elementary properties of the Fourier transform and Lemma 3.1 in the laststep. Therefore, we conclude (3.3). We proceed now to compute S(t)v0(x). Using (3.2)and (3.3), we have

S(t)v0(x) = (4πσ2)d2

ˆ

Rde2πix·ξ−4π2it|ξ|2−4π2σ2|ξ−ξ0|2dξ

LECTURES 4 AND 5 9

Completing the square, we get that the phase is given by

−4π2(σ2 + it)|ξ|2 + 2πξ · (ix+ 4πσ2ξ0)− 4π2σ2|ξ0|2

= −4π2(σ2 + it)

∣∣∣∣ξ −ix+ 4πσ2ξ0

4π(σ2 + it)

∣∣∣∣2

+|ix+ 4πσ2ξ0|2

4(σ2 + it)− 4π2σ2|ξ0|2

The last two terms of this expression, which are independent of ξ, are

|ix+ 4πσ2ξ0|24(σ2 + it)

− 4π2σ2|ξ0|2 =1

σ2 + it

(− |x|

2

4+ 2πiσ2x · ξ0 + 4π2σ4|xi0|2

)− 4π2σ2|ξ0|2

= 2πix · ξ0 +1

σ2 + it

(− |x|

2

4+ 2πtx · ξ0 − 4π2σ2it|ξ0|2

)

= 2πix · ξ0 +1

σ2 + it

(− 1

4|x− 4πtξ0|2 − 4π2it(σ2 + it)|ξ0|2

)

= 2πix · ξ0 − 4πit|ξ0|2 −|x− 4πtξ0|24(σ2 + it)

Therefore,

S(t)v0(x) = (4πσ2)d2 e

2πix·ξ0−4πit|ξ0|2− |x−4πξ0t|24(σ2+it)

ˆ

Rde−4π2(σ2+it)

∣∣ξ− ix+4πσ2ξ04π(σ2+it)

∣∣2dξ (3.4)

We would like to further simplify the above expression, in particular computing the integralon (RHS). For it, we can use the following remark.

Remark 3.2. Let G : Re z > 0 → C be defined by

G(z) :=

ˆ

Re−z|x|

2dx

where z = a+ ib ∈ C with a > 0. We have,

G(a) =

ˆ

Re−a|x|

2dx =

√π

a

ˆ

Re−π|y|dy =

√π

a

Hence, the two functions G and H : z 7→√

πz are analytic in Re z > 0 and coincide on

the positive real axis. Since a non-zero analytic function has isolated zeros, we concludethat G ≡ H in Re z > 0. Therefore,

G(z) =

ˆ

Re−z|x|

2dx =

√π

z

in Re z > 0.Now we can apply the remark to (3.4), where z := 4π2(σ2 + it). Note that z has strictly

positive real part. Thus, we getˆ

Rde−4π2(σ2+it)

∣∣ξ− ix+4πσ2ξ04π(σ2+it)

∣∣2dξ =

√π

4π2(σ2 + it)

and

S(t)v0(x) =

(σ2

σ2 + it

) d2

e2πix·ξ0−4πit|ξ0|2− |x−4πξ0t|2

4(σ2+it) .

We see that S(t)v0(x) is a localised wave with group velocity 4πξ0. We observe the followingphenomena:

• The width of the wave packet increases since Re 14(σ2+it)

< 14σ2 .


• The amplitude of the wave packet decreases since(

σ2

σ2+it

)d/2 → 0 as t→∞.

We now explore the limiting case of Remark 3.2, i.e., what happens when Re z = 0.

Remark 3.3 (Fresnel integral). The Fresnel integral satisfies the relationˆ

Re−ix

2dx =

√π

i

as we now show. Consider the contour shown in Figure 1.

R

Figure 1. Contour of integration for computing Fresnel integral.

We have

limR→∞

ˆ

γ=γ(R)e−z

2dz = ei

π4

ˆ ∞

0e−it

2dt =

√i

2

ˆ

Re−ix

2dx

and

limR→∞

ˆ

γ1

e−z2dz + lim

R→∞

ˆ

γ2

e−z2dz =

ˆ ∞

0e−x

2dx+ 0 =

√π

2

Therefore,ˆ

Re−ix

2dx =

√π

2

as we wanted to show.

We conclude this section by presenting another representation of the Schrodinger oper-ator. Consider the kernel

Kt(x) := F−1(e−4π2it|ξ|2)(x) =

ˆ

Rde−4π2it|ξ|2e2πix·ξdξ =

d∏

j=1

(e−

x2j4it

ˆ

Re−4π2it

(ξj−

xj4πt

)2dξj

)=

1

(4πit)d2

e−|x|24it

It is clear by taking Fourier transforms that S(t)u0(x) = Kt ∗ u0(x). Therefore, we have

S(t)u0(x) = Kt ∗ u0(x) =1

(4πit)d2

ˆ

Rde−|x−y|2

4it u0(y)dy

LECTURES 4 AND 5 11

Lemma 3.4 (Dispersive estimate).

||S(t)f ||L∞x (Rd) .1

|t| d2||f ||L1

x(Rd)

3.1. Dispersive relation. If we instead take a space-time Fourier transform in the linearSchrodinger equation (3.1), we get

−2πτu(τ, ξ)− 4π2|ξ|2u(τ, ξ) = −2π(τ + 2π|ξ|2)u(τ, ξ) = 0

i.e., the space-time Fourier transform u(τ, ξ) of a solution to the linear Schrodinger equationis a distribution supported on the paraboloid τ + 2π|ξ|2 = 0. Therefore, the dispersiverelation for the linear Schrodinger equation is given by

ω(ξ) = 2π|ξ|2

with

• phase velocity ω(ξ) ξ|ξ|2 = 2πξ

• group velocity ∇ω(ξ) = 4πξ

LECTURES 6 - 8

REUBEN WHEELER

1. Lecture 6

We have the following two estimates. The first is called the dispersive estimate and thesecond is the statement of the solution map being unitary on L2.

‖S(t)f‖L∞x. 1

|t| d2‖f‖L1

x(1.1)

‖S(t)f‖L2x

= ‖f‖L2x

(1.2)

By application of Riesz-Thorin interpolation between these two estimates we obtain thefollowing estimate,

‖S(t)f‖Lpx. 1

|t|d(12− 1

p)‖f‖

Lp′x. (1.3)

As θ satisfying the relations 1p′ = θ

1 + 1−θ2 , 1

p = θ∞ + 1−θ

2 . In particular 1p + 1

p′ = 1, p ∈ [2,∞]

and θ2 = 1

2 − 1p is given by 1− 2

p .

Definition 1.1. We say (q,r) is a Schrodinger admissible if 2 ≤ q, r ≤ ∞, (q, r, d) 6=(2,∞, 2)

2

q+d

r=d

2(1.4)

(Scaling Condition) Note that these conditions describe a line segment in the (1q ,1r )-plane.

Proposition 1.2. (Strichartz Estimates)

(1) (Homogeneous Estimate) For admissible (q,r)

‖S(t)f‖LqtL

rx(R×Rd) . ‖f‖L2

x, ∀f ∈ L2

x (1.5)

(2) (Dual Homogeneous Estimate)∥∥∥∥ˆ

RS(−t′)F (t′)dt′

∥∥∥∥L2(Rd)

. ‖F‖Lq′Lr′x

(1.6)

(3) (Retarded Estimate/Nonhomogeneous Estimate) For admissible(q, r)

‖ˆ t

0S(t− t′)F (t′)dt′‖Lq

tLrx. ‖F‖

Lq′t L

r′x

(1.7)

Remark 1.3. In the time-averaged sense (Lqt ), S(t)f gains integrability because S(t)f ∈ Lrxfor a.e. t. This is an effect where high peaks are smoothed out.

Proof. (non-endpoint case, proof of 1,2) The proof follows a TT ∗ argument. With anoperator T : H → B, T ∗ : B′ → H, TT ∗ : B′ → B where H is a Hilbert space, B is aBanach space we have that

‖T‖ <∞ ≡ ‖T ∗‖ <∞ ≡ ‖TT ∗‖ <∞ (1.8)1

2 REUBEN WHEELER

We let T = S(t), we have that

〈S(t)f,G(t, x)〉t,x =

ˆ ˆ

S(t)fG(t, x)dtdx (1.9)

ˆ

f(x)

ˆ

S(−t)G(t, x)dtdx = 〈f,ˆ

S(−t)G(t, x)dt〉,x (1.10)

so that T ∗G(x) =´

S(−t)G(t, x)dt. TT ∗F = S(t)´

R S(−t′)F (t′)dt′ =´

R S(t− t′)F (t′)dt′.

Lemma 1.4. (Hardy-Littlewood-Sobolev Inequality) 1 < p, q, r <∞, 1r + 1 = 1p + 1

q

‖ 1

|x|np

∗ f‖Lrx(Rn) . ‖f‖Lq

x(Rn) (1.11)

Remark 1.5. This inequality is a near-miss for Young’s inequality, application of Young’sinequality would give

LHS . ‖f‖Lq‖ 1

|x|dp

‖Lpx

(1.12)

where the second term in the product is infinite.

Remark 1.6. H-L-S ⇐⇒ Sobolev Inequality:

‖f‖Lr . ‖f‖W s,p (1.13)

⇐⇒ ‖|∇|−sf‖Lr = ‖F−1(|ξ|−sf)‖Lr = ‖|x|d−s ∗ f‖Lr . ‖f‖qL (1.14)

Applying Minkowski’s inequality, the inequality 1.3, and H-L-S successively we obtainthe following sequence of inequalities for r ≥ 2

‖TT ∗F‖LqtL

rx≤ ‖

ˆ

‖S(t− t′)F (t′)‖Lrxdt′‖Lq

t(1.15)

. ‖ˆ

1

|t− t′|d( 12− 1r)‖F (t′)‖Lr′

xdt′‖Lq

t(1.16)

. ‖F‖Lq′t L

r′x

(1.17)

This proves 1, 2 by the TT ∗ argument. The H-L-S condition in one dimension is verified1q + 1 = d(12 − 1

r ) + 1q′ ≡ 2

q + dr = d

2 which is the condition for Schrodinger admissibility.

The further condition is

0 <1

q, d(

1

2− 1

r),

1

q′< 1 (1.18)

≡ 1 < q <∞, 2 < r <2d

d− 2(1.19)

For the endpoint cases we have r = 2, q = ∞, where 1 follows by unitarity of S(t). Forq = 2, r = 2d

d−2 , d ≥ 3 see Keel-Tao American Journal of Mathematics, Volume 120, Number

5, October 1998 pp. 955-980 (Article)

LECTURES 6 - 8 3

2. Lecture 7

It remains to prove 3. Again making use of 1.3 and H-L-S as in the previous proof wehave that

‖ˆ t


tLrx. ‖F‖

Lq′t L

r′x

(2.1)

for (q, r), (q, r) Schrodinger admissible

Proof. (q, r) = (q, r)

‖ˆ t


tLrx

= ‖ˆ

RS(t− t′)χ[0,t](t

′)F (t′)dt′‖LqtL

rx

(2.2)

. ‖ˆ

R

1

|t− t′|d(12− 1

p)‖χ[0,t](t

′)‖Lr′xdt′‖Lq

t(2.3)

. ‖F‖Lq′t L

r′x

(2.4)

The endpoint cases we consider are given by

(1)

‖ˆ

t<t′S(t− t′)F (t′)dt′‖Lq

tLrx. ‖F‖

Lq′t L

r′x

(2.5)

(2)

‖ˆ

t<t′S(t− t′)F (t′)dt′‖L∞

t L2x. ‖F‖

Lq′t L

r′x

(2.6)

If we have a result of the form of 1 the desired result follows as χ(0,t] = χ(−∞,t]−χ(−∞,0].We turn to the proof of 2. We have that

|ˆ

t′<tS(t− t′)F (t′)dt′|2 =

⟨ˆ

t1<tS(t− t1)F (t1)dt1,

ˆ

t2<tS(t− t2)F (t2)dt2

⟩(2.7)

=

⟨ˆ

t1<tF (t1)dt1,

ˆ

t2<tS(t1 − t)S(t− t2)F (t2)dt2

⟩(2.8)

=

ˆ

t1<t

⟨(t1),

ˆ

t2<tS(t1 − t2)F (t2)dt2

⟩dt1 (2.9)

≤ˆ

R‖F (t1)‖Lr′

x

∥∥∥∥ˆ

t2<tS(t1 − t2)F (t2)dt2

∥∥∥∥Lrx

dt1 (2.10)

≤ˆ

R‖F‖

Lq′t L

r′x

∥∥∥∥ˆ

t2<tS(t1 − t2)F (t2)dt2

∥∥∥∥LqtL

rx

(2.11)

. ‖F‖2Lq′t L

r′x

(2.12)

where the inequalities are obtained by the successive application of Holder’s inequality inx, Holder’s inequality in t and finally as a consequence of 1.

We want to show 3ˆ ˆ

t′<t〈S(t− t′)F (t′), G(t)〉L2

xdt′dt . ‖F‖

Lq′t L

r′x‖G‖

Lq′t L

r′x. (2.13)

Fix (q, r). We know that 2.13 holds for q = q, r = r and also for q = ∞, r = 2 as aconsequence of 1 and 2, respectively. Interpolating between these shows that 2.13 holds forall q ≥ q (and r = r(q)).

4 REUBEN WHEELER

The remaining case of q < q follows by a duality argument. With q ≥ q we find∣∣∣∣∣

ˆ

t′

⟨F (t′),

ˆ

t>t′S(t′ − t)G(t)dt

⟩

L2x

dt′∣∣∣∣∣ (2.14)

. ‖F‖Lq′t L

r′x‖G‖

Lq′t L

r′x

(2.15)

≡ ‖F‖Lq′t L

r′x‖G‖

Lq′t L

r′x

(2.16)

By duality this is equivalent to

‖ˆ

t>t′S(t′ − t)G(t)dt‖

LqtL

rx. ‖G‖

Lq′t L

r′x

(2.17)

By relabelling we have

‖ˆ

t>t′S(t′ − t)G(t)dt‖Lq

tLr . ‖G‖

Lq′t L

r′x

(2.18)

, for q ≤ q.Since χ[−∞,t](t′) = χR(t′)− χ(t,∞)(t

′) 2.13 holds for q ≤ q.

For more information related to the Christ-Kiselev lemma (used for the insertion of χ[0,t])see Tao’s book ”Nonlinear dispersive equations: local and global analysis”.

2.1. Local Well Posedness of Nonlinear Schrodinger Equation. For d = 1, p = 3we have that scrit = d

2 − 2p−1 = −1

2 . This is the cubic NLS on R.

Theorem 2.1. The cubic NLS on R is locally well-posed in Hs(R).

Proof. s = 0. Let u0 ∈ L2(R). On [−T, T ] (NLS)⇔ u(t)Γu0(u)(t) = S(t)u0 ∓ i´ t0 S(t −

t′)χ[−T,T ](t′)|u|2u(t′)dt′ Note: (q, r) = (8, 4) is admissable as 28 + 1

4 = 12 . We write XT =

C([−T, T ];L2x(R))∩L8([−T, T ];L4

x) = CTL2x∩L8

TL4x equipped with the norm ‖u‖V +‖u‖W

where X = V ∩W .

‖Γu0(u)‖XT≤ C1‖u0‖L2

x+ C2‖|u|2u‖

L87T L

43x

(2.19)

≤ C1‖u0‖L2x

+ C3T12 ‖|u|2u‖

L83T L

43x

(2.20)

≤ C1‖u0‖L2x

+ C3T12 ‖u‖L8

TL4x≤ C1‖u0‖L2

x+ C3T

12 ‖u‖3XT

(2.21)

Here we have applied the Strichartz estimates, and Holder’s inequality in T and x. Thuswe have, for u ∈ BR ⊂ XT ,

‖Γu0(u)‖XT≤ C1‖u0‖L2

x+ C3T

12R2R. (2.22)

Choosing R = ‖u0‖L2x

and T = T (R) such that C3T12R2 < C1 we have obtained the

inequality

‖Γu0(u)‖XT≤ 2T

12C1‖u0‖L2

x(2.23)

LECTURES 6 - 8 5

Similarly, for u, v ∈ XT

‖Γu0(u)− Γu0(v)‖XT≤ C3T

12 ‖|u|2u− |v|2v‖

L83T L

43x

(2.24)

which is a telescoping sum uuu−vvv = (u−v)uu+v(u− v)u+vv(u−v). Applying Young’sinequality we have the bound

C4T12(‖u‖2XT

+ ‖v‖2XT

)‖u− v‖XT

(2.25)

≤ 2C4T12R2 ≤ 1

2(2.26)

By choosing T12 ∼ R−2 i.e. T ∼ R−4, we conclude that Γu0 is a contraction on BR ⊂ XT ,

R = 2C1‖u0‖L2x. Therefore, by the Banach fixed point theorem, there exists a unique

u ∈ BR ⊂ XT such that u = Γu0(u).

Remark 2.2. (1) local existence time T = T (‖u0‖L2) ∼ 1‖u0‖4

L2. So that, together with

conservation of the L2 norm, global well-posedness follows from local well-posednessin L2(R) by a bootstrap argument.

(2) Uniqueness holds in BR ⊂ CTL2x ∩ L8

TL4x, which we call conditional uniqueness in

CTL2x ∩ L8

TL4x

(3) We should also show continuity of u in t (taking values in L2).

3. Lecture 8

The 1-dimensional cubic nonlinear Schrodinger equation is given by

i∂tu+ ∆u = ±|u|2u, x ∈ R (3.1)

In this lecture we prove that global well-posedness of the cubic NLS follows from localwell-posedness by mass conservation.

What if u0 ∈ Hs(R) for some s > 0? If s ∈ N , we can use the Leibniz rule

∂x(fg) = ∂xf · g + f · ∂xg. (3.2)

For general s > 0, we require use of the fractional Leibniz rule which is stated as follows.For s ∈ (0, 1], 1 < r, p1, p2, q1, q2 <∞ such that 1

r = 1pj

+ 1qj

for j = 1, 2

‖|∇|s(fg)‖Lrx. ‖|∇|sf‖Lp1‖g‖Lq1 + ‖f‖Lp2‖|∇|sg‖Lq2 (3.3)

Remark 3.1. Morally this may be thought of as |∇|s(fg) ≈ (|∇|sf)g + f(|∇|sg). Fors = 0 the result follows from Holder’s inequality. For s = 1 the result can be viewed as anapplication of the ”Leibniz rule” and Holder’s inequality (though strictly we require s ∈ 2Nto apply the Leibniz rule).

The Leibniz rule for fractional integration, together with the relation 〈ξ〉s ∼ 1 + |ξ|simplies the following

‖〈∇〉s(fg)‖Lrx. ‖〈∇〉sf‖Lp1‖g‖Lq1 + ‖f‖Lp2‖〈∇〉sg‖Lq2 (3.4)

We now use this result to prove local well-posedness of the cubic NLS in Hs(R), s ≥ 0.

Given T > 0, we define XsT = CTH

s ∩ L8TW

s,4x .

‖Γu0‖XsT

= ‖〈∇〉su0‖L2x

+ ‖〈∇〉s(|u|2u)‖L

87T L

43x

(3.5)

. ‖u0‖Hs + T12 ‖〈∇〉s(|u|2u)‖

L87T L

43x

(3.6)

6 REUBEN WHEELER

. ‖u0‖Hs + T12 ‖u‖3

L8TL

s,4x. (3.7)

Where we first applied Strichartz estimates, and then Holder in T and inequality 3.4. Thefactor of ‖u‖3

L8TL

s,4x. ‖u‖3Xs

T. Applying the same estimates and a telescoping sum we have

‖Γu0(u)− Γu0(v)‖XsT

= ‖〈∇〉s(|u− v|2(u− v))‖L

87T L

43x

(3.8)

. T 12

(‖u‖2Xs

T+ ‖v‖2Xs

T

)‖u− v‖Xs

T. (3.9)

If we choose T = T (‖u0‖Hs) > 0 sufficiently small, Γu0 is a contration on BR ⊂ XsT

where R ∼ ‖u0‖Hs . By Banach’s fixed point theorem this proves local well-posedness ofNLS in Hs(R) for s ≥ 0.

Together with mass conservation and persistence of regularity this implies global well-posedness in Hs(R) for s ≥ 0. It remains to prove persistence of regularity

‖〈∇〉s(|u|2u)‖L

83T L

43x

. ‖u‖L8TW

s,4x‖u‖2L8

TL4x. (3.10)

For d = 2, p = 3 we have that scrit = d2 − 2

p−1 = 0.

Theorem 3.2. The Cubic NLS on R2 is locally well-posed in L2(R2) and is also globallywell-posed in L2(R2) with small initial data (‖u0‖L2 is sufficiently small).

(q, r) = (4, 4) and (q, r) = (∞, 2) are Schrodinger admissable (2q + dr + d

2). XT = X0T =

CTL2 ∩ L4

TL4x.

‖Γu0(u)‖XT≤ ‖S(t)u0‖XT

+ C2‖|u|2u‖L

43T L

43x

≤ C1‖u0‖L2 + C2‖u‖3L4TL

4x

(3.11)

‖Γu0(u)− Γu0(v)‖XT≤ C2‖|u|2u− |v|2v‖

L43T L

43x

≤ C3

(‖u‖2L4

TL4x

+ ‖v‖2L4TL

4x

)‖u− v‖L4

TL4x.

(3.12)We now try to bound the coefficient of ‖u − v‖L4

TL4x

in order for the operator to be a

contraction. We set R = 2C1‖u0‖L2 and let u ∈ BR ⊂ XT . Then we have that

‖Γu0(u)‖XT≤ 1

2R+ C2R

3 ≤ R, (3.13)

provided C2R2 ≤ 1

2 .

‖Γu0(u)− Γu0(v)‖XT≤ 2C3R

2‖u− v‖XT≤ 1

2‖u− v‖XT

(3.14)

provided C3R2 ≤ 1

4 . These estimates hold true also for T = ∞ so that, for small initial

data in L2(R2) we obtain global well-posedness by a fixed point argument in BR.What about large L2-data?‖S(t)u0‖L4(Rt;L4

x). ‖u0‖L2 by a Strichartz estimate. It follows from the dominated

convergence theorem that

limT→∞

‖S(t)u0‖L4TL

4x

= 0 (3.15)

but ‖S(t)u0‖L∞T L2

x= ‖u0‖L2

x9 0 as T →∞.

‖Γu0(u)‖L4TL

4x≤ ‖S(t)u0‖L4

TL4x

+ C2‖|u|2u‖L

43T L

43x

(3.16)

LECTURES 6 - 8 7

Fix η > 0 small. Then, given u0 ∈ L2(R2), there exists T = T (u0) > 0 small such that‖S(t)u0‖L4

TL4x≤ 1

2η. Let u ∈ Bη ⊂ L4TL

4x. Then we have that

‖Γu0(u)‖L4TL

4x≤ 1

2+ C2η

3 ≤ η (3.17)

Take u ∈ Bη ⊂ L4TL

4x. We find that

‖Γu0(u)‖L4TL

4x≤ C3

(‖u‖2L4

TL4x

+ ‖v‖2L4TL

4x

)‖u− v‖L4

TL4x≤ 2C3η

2‖u− v‖L4TL

4x,∀u, v ∈ Bη

(3.18)It now follows from the contraction mapping principle that there exists a unique u ∈

Bη ⊂ L4TL

4x which is a solution to the non-linear Schrodinger equation.

By 3.11,‖u‖CTL2 = ‖Γu0(u)‖CTL2 ≤ C1‖u0‖L2 + C2η

3 <∞ (3.19)

so that u ∈ CTL2 ∩ L4TL

4x. Since T = T (u0) depends on ”the profile of u0” this indicates

the critical nature of the problem.We could simply run a contraction argument in

AR,η = u ∈ BR ⊂ XT , andu ∈ Bη ⊂ L4TL

4x (3.20)

with R ∼ ‖u0‖L2 , η << 1, endowed with the XT -norm. 3.11 =⇒ ‖Γu0(u)‖XT≤

C1‖u0‖L2 + C2η3 ≤ 2C1‖u0‖L2 = R 3.12 =⇒ ‖Γu0(u)− Γu0(v)‖XT

≤ 2C3η2‖u− v‖XT

Also, ‖Γu0(u)‖L4TL

4x≤ 1

2η + η3 ≤ η.

.

DISPERSIVE PDE, LECTURES 9-10

FELIX JAMES-KAHN

1. More on estimates

1.1. Dispersive estimate for the Schrodinger equation. Previously we proved thedispersive estimate for the Schrodinger equation. Namely that if u satisfies:

i∂tu = ∆u

u|t=0 = f,(t, x) ∈ R+ × Rd, (1.1)

then we have the following, which is called the dispersive estimate:

‖u‖L∞x . |t|−d/2‖f‖L1

We proved this by using the solution operator to write u = S(t)f = Kt ∗ f for somekernel Kt. Calculating this kernel explicitly gave:

Kt(x) = (4πi|t|)−d/2e−|x|24it

However, we would like to be able to complete the proof without computing such anexpression for a kernel. This will be the goal for this section.

We begin with the case d = 1. Using the solution operator S(t) we see that we can writeKt as the following integral:

Kt(x) = F−1(e−itξ2) =

ˆ

Re−itξ

2+ixξdξ

It suffices to prove:

Proposition 1.1.

‖Kt‖L∞ . |t|−1/2

since then by Young’s inequality we have:

‖u‖L∞ = ‖Kt ∗ f‖L∞ ≤ ‖Kt‖L∞‖f‖L1 . |t|−1/2‖f‖L1

Proof. Taking t > 0 if we perform the substitution ζ = t1/2ξ we see that:

Kt(x) =1√t

ˆ

Re−iζ2+i x√

tζdζ =

1√tK1(y)

where y = x/√t. Thus:

‖Kt‖L∞x =1√t‖K1‖L∞y

so it remains to show that K1 is bounded.Our main tool will be the method of stationary phase. Fixing x, we will write K1 as

follows:1

2 FELIX JAMES-KAHN

K1(x) =

ˆ

Re−iξ

2+ixξdξ =

ˆ

Re−iφ(ξ)dξ

where φ(ξ) = ξ2 − xξ. The idea is to make use of integration by parts. Since

e−iφ(ξ) =∂ξe−iφ(ξ)

−iφ′(ξ)we use integration by parts to perform manipulations such as:

ˆ

e−iφ(ξ)dξ =

ˆ

∂ξ

(e−iφ(ξ)

) 1

−iφ′(ξ)dξ =

ˆ

e−iφ(ξ)∂ξ

(1

−iφ′(ξ)

)dξ

Note that the presence of φ′(ξ) = 2ξ − x in the denominator of these fractions will helpus if it is not small. To make use this we introduce a smooth cutoff function.

Let ψ ∈ C∞(R; [0, 1]) such that ψ(x) = 0 if |x| ≥ 2 and ψ(x) = 1 if |x| ≤ 1. We nowwrite our integral as

K1(x) =

ˆ

Re−iφ(ξ)ψ(2ξ − x)dξ +

ˆ

Re−iφ(ξ)(1− ψ)(2ξ − x)dξ =: I(x) + II(x)

A bound for I follows immediately since |ψ| ≤ 1 and is supported in [−2, 2]. So:

|I(x)| ≤ˆ

|ψ(2ξ − x)|dξ . 1

To deal with II we will use integration by parts:

|II(x)| =ˆ

e−iφ(ξ)∂ξ

((1− ψ)(2ξ − x)dξ

−iφ′(ξ)

)dξ

=

ˆ

e−iφ(ξ)(−2

(1− ψ)(2ξ − x)

i(2ξ − x)2

)dξ +

ˆ

e−iφ(ξ)(1− ψ)′(2ξ − x)

φ′(ξ)dξ

For the first term we note that |(1 − ψ)(2ξ − x)| ≤ 1 and is supported on |2ξ − x| ≥ 1.For the second term we note that (1 − ψ)′ = −ψ′ which is supported on [−2,−1] ∪ [1, 2].Thus (1− ψ)′(2ξ − x) is supported on 2ξ − x ∈ [−2,−1] ∪ [1, 2]. In particular, |φ′(ξ)| ≥ 1.Thus:

|II(x)| ≤ Cˆ

|2ξ−x|≥1(2ξ − x)−2dξ +

ˆ

2ξ−x∈[−2,−1]∪[1,2]|(1− ψ)′(2ξ − x)|dξ . 1 + 1

So |K1(x)| . 1 for any x.

1.2. Glimpse of oscillatory integrals. The method of stationary phase is a method forbounding the size of so called oscillatory integral. In this section we study these integralsand discuss other methods. For λ > 0 we are interested in integrals of the form:

I(λ) =

ˆ b

ae−iλφ(x)ψ(x)dx

where the phase function φ is real valued, and ψ is a complex valued smooth compactlysupported cutoff function. We are interested in controlling the size of |I(λ)| in terms of λ.

DISPERSIVE PDE, LECTURES 9-10 3

Lemma 1.2. Suppose that ψ has compact support contained in (a, b), and that φ′(x) ≥ cfor some c > 0 on the support of ψ. Then for all N ∈ N

|I(λ)| . λ−N

where the implicit constant may depend on N and the CN norms of ψ and ψ.

Proof. We define the differential operator:

Df :=1

iλφ′df

dx

Note that on the support of ψ we have that φ′ 6= 0 so this is well defined. Also D(e−iλφ) =e−iλφ and the transpose operator is:

DT f = − d

dx

(f

iλφ′

)

. Now let N ∈ N. By using integration by parts N times we get:

I(λ) =

ˆ b

aDN (e−iλφ)ψ =

ˆ b

ae−iλφ(DT )Nψ

Note that there are no boundary terms from the integration by parts since we haveassumed ψ is supported inside (a, b) (and so vanishes at the endpoints). It this remains tobound |(DT )Nψ|. However, each time we apply DT to ψ we can pull about a factor of 1/λgiving a factor of λ−N in total; and the remaining terms will, by the quotient rule, be acombination of the derivatives of ψ and φ divided by φ′2N . The former can be bounded bythe CN norms and that latter is controlled since we have assumed |φ′| ≥ c. Thus we have

|I(λ)| . λ−N

as required.

Proposition 1.3 (van der Corput). Suppose φ is smooth on (a, b) and there exists some

k ∈ N such that |φ(k)| ≥ 1 on [a, b]. Then provided either:

(1) k ≥ 2 or;(2) k = 1 and φ′ is monotone.

Then:

∣∣∣∣ˆ b

aeiλφ

∣∣∣∣ ≤ ckλ−1/k

where ck is a constant depending on k (but independent of a and b)

Proof. We start by assuming condition (2) where k = 1. Using a differential operator Ddefined similarly to the one defined in the proof of the previous proposition, we have:

ˆ b

aeiλφdx =

ˆ b

aD(eiλφ) · 1dx =

ˆ b

aeiλφDT (1)dx+

eiλφ

iλφ′

∣∣∣b

a

Taking the absolute value, we can immediately bound the boundary term by 2/λ sinceφ′ ≥ 1 on [a, b]. We turn to the remaining term:

4 FELIX JAMES-KAHN

∣∣∣∣ˆ b

aeiλφDT (1)dx

∣∣∣∣ =1

λ

∣∣∣∣ˆ b

aeiλφ

d

dx

(1

φ′

)dx

∣∣∣∣ ≤1

λ

ˆ b

a

∣∣∣∣d

dx

(1

φ′

)∣∣∣∣ dx

Now we use the monotonicity of φ′ to bring the absolute values outside the integral again:

=1

λ

∣∣∣∣ˆ b

a

d

dx

(1

φ′

)dx

∣∣∣∣ =1

λ

∣∣∣∣1

φ′(b)− 1

φ′(a)

∣∣∣∣ . λ−1

which is the desired result in this case.We move on to the case (1) where k ≥ 2. We proceed by induction on k. Suppose the

result holds for k. Now without loss of generality we assume φ(k+1)(x) ≥ 1 for all x ∈ [a, b].

Let c ∈ [a, b] be the (unique) point x = c such that φ(k)(x) attains its minimum.

If φ(k)(c) = 0 then φ(k)(x) ≥ δ on (c − δ, c + δ)c (since its derivative is at least 1). Wethus split up the integral as:

ˆ b

a=

ˆ c−δ

a+

ˆ c+δ

c−δ+

ˆ b

c−δWe have that

∣∣∣∣ˆ c+δ

c−δeiλφ

∣∣∣∣ ≤ 2δ

.On the other hand, by writing iλφ as i(λδ)(φ/δ) we have that φ′/δ ≥ 1 and so applying

our inductive hypothesis gives:

∣∣∣∣ˆ c−δ

aeiλφ

∣∣∣∣ . ck(λδ)−1/k

and similarly for the integral´ bc+δ . By choosing δ = λ−

1k+1 the bounds of all terms agree

with the desired bound.If φ(k)(c) 6= 0 then c = a or c = b. In this case we proceed as above, but we only have to

split the integral into two parts, eg,´ a+δa +

´ ba+δ.

Corollary 1.4. Let ψ be smooth and of compact support. Under the same assumptions onφ as above we have:

∣∣∣∣ˆ b

aeiλφψ

∣∣∣∣ ≤ ckλ−1/k(|ψ(b)|+

ˆ b

a|ψ′|)

Proof. Let F (x) =´ xa e

iλφ(y)dy. Thus, using the fundamental theorem of calculus andintegrating by parts:

ˆ b

aeiλφ(x)ψ(x)dx =

ˆ b

aF ′(x)ψ(x)dx = F (b)ψ(b)− F (a)ψ(a)−

ˆ b

aF (x)ψ′(x)dx

Note that F (a) = 0. Now taking the absolute value and using the fact that by the

proposition, |F (x)| ≤ ckλ−1/k for all x ∈ [a, b], we get the desired bound.

Lecture 10:


1.3. A glimpse of the Littlewood-Paley decomposition. The purpose of theLittlewood-Paley decomposition is to isolate the behaviours that occur at different fre-quencies (regions of ξ-space). Let us introduce a smooth bump function φ that satisfiesφ(ξ) = 1 for |ξ| ≤ 1, and φ(ξ) = 0 for |ξ| > 2. So φ is supported in a ball of radius 2 infrequency space.

Now let N be dyadic number; ie, a number of the form N = 2i for some i ∈ Z. Wedefine:

P≤Nf = F−1(φ(ξ/N)f(ξ))

Note that φ(ξ/N) satisfies φ(ξ) = 1 for |ξ| ≤ N , and φ(ξ) = 0 for |ξ| > 2N . Thus we callthis the Littlewood-Paley projection onto frequencies . N . Similarly we define

PNf = P≤Nf − P≤N/2fHere the cutoff function on the Fourier side is φ(ξ/N)− φ(2ξ/N) which is supported in

the annulus N/2 ≤ |ξ| ≤ 2N . We call PN the Littlewood-Paley projection onto frequencies∼ N .

It is possible to recover f from its Littlewood-Paley projections:

• P≤Nf =∑

dyadic M≤N PMf• f = limN→∞ P≤Nf• f =

∑M dyadic PMf

Theorem 1 (Littlewood-Paley theory). If 1 < p <∞ then:

‖f‖Lp ∼

∥∥∥∥∥∥∥

∑

N dyadic

|PNf |2

1/2∥∥∥∥∥∥∥Lp

We will not prove this theorem, but note that for p = 2 the result follows immediatelyfrom Plancherel’s theorem. The function:

∑

N dyadic

|PNf |2

1/2

is called the square function of f .

Theorem 2 (Bernstein’s inequalities). Let 1 ≤ p ≤ q ≤ ∞. Then:

(1) For s > 0, ‖P≤N |∇|sf‖Lp . N s‖P≤Nf‖Lp

(2) For s ∈ R, ‖PN |∇|sf‖Lp ∼ N s‖PNf‖Lp

(3) ‖P≤Nf‖Lq . Ndp− d

q ‖P≤Nf‖Lp

(4) ‖PNf‖Lq . Ndp− d

q ‖PNf‖Lp

We do not prove inequalities (3), (4) but note that they can be thought of as frequencylocalised analogues of the Sobolev inequality with s = d

p − dq .

Proof. We begin with (2). Define:

PN := PN/2 + PN + P2N = P≤2N − P≤N/4

6 FELIX JAMES-KAHN

which is the projection onto the wider range of frequencies N/4 ≤ |ξ| ≤ 4N . This willbe useful since our PN are not true projections in the sense that PN 6= PNPN . However

we do have PN = PNPN , and PN has essentially the same properties as PN : Since the

supports of at most three PN overlap at once we have∑

N PNf = 3f and in particular theLittlewood-Paley theorem still holds, ie:

‖f‖Lp ∼

∥∥∥∥∥∥∥

∑

N dyadic

|PNf |2

1/2∥∥∥∥∥∥∥Lp

Since our operators are Fourier multipliers they commute, so we have:

PN |∇|sf = |∇|sPNf = |∇|sPNPNfWe consider the Fourier multiplier of the operator |∇|sPN . It is N s|ξ|sψ(ξ/N) where

ψ(ξ) = φ(ξ/2)−φ(4ξ). Putting aside the factor ofN s for now, we consider the correspondingconvolution kernel for N = 1:

Ks = F−1(|ξ|sψ(ξ))

Since ψ(ξ) is supported away from 0, and |ξ|s is smooth everywhere apart from 0, wehave that |ξ|sψ(ξ) is smooth. Thus its (inverse) Fourier transform must have rapid decayand in particular ‖Ks‖L1 <∞. Now we consider the kernels for other N .

F−1(|ξ/N |sψ(ξ/N)) = F−1(Ks(ξ/N)) = NdKs(Nx) =: KN,s(x)

and we note that:

‖KN,s‖L1 = Nd

ˆ

|Ks(Nx)|dx = ‖Ks‖L1 <∞

is a bound independent of N . Now proceeding using Young’s inequality:

‖PN |∇|sf‖Lp = N s‖KN,s ∗ (PNf)‖Lp ≤ N s‖KN,s‖L1‖PNf‖Lp . N s‖PNf‖Lp

as required. We can prove the reverse inequality by arguing in a similar way; alternatively,if g = |∇|sf then f = |∇|−sg. Now by what we proved:

‖PNf‖Lp = ‖PN |∇|−sg‖Lp . N−s‖PNg‖Lp = N−s‖PN |∇|sf‖Lp

as required.We now turn to (1). This follows from (2), though it can also be proved directly in a

similar way.Since:

P≤N |∇|sf =∑

M≤NPM |∇|sf

we have by Minkowski’s inequality and (2):

‖P≤N |∇|sf‖Lp ≤∑

M≤N‖PM |∇|sf‖Lp .

∑

M≤NM s‖PMf‖Lp

Note that since we are summing over dyadic M we have:


∑

M≤NM s =

∞∑

j=0

N2−js

= N s∑

2−js ∼ N s

since s > 0. Thus continuing we have:

‖P≤N |∇|sf‖Lp . N s supM≤N

‖PMf‖Lp ≤ N s

∥∥∥∥∥∥∥

∑

M≤N|PMf |2

1/2∥∥∥∥∥∥∥Lp

by Minkowski’s (integral) inequality. Now since we take 1 < p <∞ the result follows bythe Littlewood-Paley theorem. 1.4. Return to the dispersive estimate. Previously we proved the dispersive estimatefor d = 1. Here we treat the higher dimensional cases. We wish to show that the estimateholds for PNf , ie, for each dyadic N :

‖S(t)PNf‖L∞x (Rd) ≤C

|t|d/2 ‖PNf‖L1x(Rd) (1.2)

This will be sufficient for proving the Strichartz estimate. Indeed, by the unitarity of S(t):

‖S(t)PNf‖L2 = ‖PN‖L2

and interpolating this with (1.2) gives:

‖S(t)PNf‖Lpx. 1

|t|d(12− 1

p)‖PNf‖Lp′

x(1.3)

for 2 ≤ p ≤ ∞. In particular, by the Littlewood-Paley theorem, for 2 ≤ p <∞ we have:

‖S(t)f‖Lpx∼∥∥∥(∑

N |S(t)PNf |2)1/2∥∥∥

Lp= ‖S(t)PNf‖Lp

x`2N

Now by Minkowski’s inequality (since p ≥ 2) and then (1.3):

≤ ‖S(t)P )nf‖`2NLpx. 1

|t|d(12− 1

p)‖PNf‖`2NLp′

x

Now since p′ ≤ 2 we can apply Minkowski’s inequality again whence one final use of theLittlewood Paley theorem gives us the Strichartz estimate:

≤ 1

|t|d(12− 1

p)‖PNf‖Lp′

x `2N

∼ 1

|t|d(12− 1

p)‖f‖

Lp′x

So it remains to prove (1.2). Let us write

S(t)PNf = S(t)PNPNf = PNS(t)(PNf) = (PNKt) ∗ (PNf)

Where Kt is the kernel of S(t) such that S(t)f = Kt ∗ f :

Kt(x) =

ˆ

Rd

ei(−4π|ξ|2t+2πix·ξ)dξ

So by Young’s inequality:

‖S(t)PNf‖L∞x ≤ ‖PNKt‖L∞x ‖PNf‖L1x

So we have reduced the problem to proving: ‖PNKt‖L∞x . |t|−d/2

DISPERSIVE PDE LECTURE 11 AND 12

FINLAY DUPREE MCINTYRE

1. Lecture 11

The goal of this lecture is to prove the following estimate for all dyadic N and t ∈ R:

‖PNKt‖L∞x .1

|t| d2, (1.1)

with the implicit constant independent of t and Kt the kernel defined by

Kt(x) =

ˆ

Rdei(−4π2|ξ|2t+2πx·ξ)dξ.

For simplicity we will drop the · and simply write P , this is not to be confused with theLittlewood-Paley projection defined in the last lecture which we will no longer need. Letus define a smooth bump function supported in the annulus |ξ| ∼ 1 by

ψ(ξ) := ϕ(ξ)− ϕ(2ξ),

note that ψ(ξ) = ψ(|ξ|) is radial. In order to prove (1.1) we consider two separate cases.

Case 1: t . N−2. By a change of variables ξ 7→ t−12 ξ we get

|PnKt(x)| = 1

|t| d2

∣∣∣∣∣

ˆ

Rdei(−4π2|ξ|2t+2πt−

12 x·ξ)ψ

(ξ

t12N

)dξ

∣∣∣∣∣.

Since φ(

ξ

t12N

)is supported in ξ ∼ t 1

2N and t12N . 1 by assumption we get

|PnKt(x)| . 1

|t| d2.

Case 2: t N−2. Here we consider two separate subcases.

Subcase 2a: |x| tN . By a change of variables ξ 7→ ξN we can write

PnKt(x) = Nd

ˆ

Rde2πiNx·ξe−4π2itN2|ξ|2ψ (|ξ|) dξ.

Writing things in polar coordinates we then get

PNkt(x) = Nd

ˆ

Sd

ˆ ∞

0eitN

2φ(r)ψ (r) rd−1drdσ(ω),

where we define φ(r) = 2πr xtN · ω − 4π2r2. Observe that

|φ′(r)| = |2π x

tN· ω − 8π2r| ∼ 1

1

2 FINLAY DUPREE MCINTYRE

in the region where r ∼ 1, since |x| tN by assumption. Note that ψ(r) = ψ(r)rd−1 is alsoa smooth function with compact support in r ∼ 1. Observe that if we are differentiatingwith respect to r we have

(eiN

2tφ(r)

iN2tφ′(r)

)′= eiN

2tφ(r) − φ′′(r)eiN2tφ(r)

iN2t(φ′(r))2

for all t ∈ R. Integration by parts gives us

ˆ ∞

0

(eiN

2tφ(r)

iN2tφ′(r)

)′ψ(r)dr = −

ˆ ∞

0

eiN2tφ(r)

iN2tφ′(r)ψ′(r)dr,

where we use the fact that ψ has compact support so the boundary term is equal to zero.So combining the last two statements we can write

ˆ ∞

0eiN

2tφ(r)ψ(r)dr = − 1

iN2t

ˆ ∞

0eiN

2tφ(r)

(φ′(r)ψ′(r) + φ′′(r)ψ(r)

(φ′(r))2

)dr.

Define a new function

ψ1(r) :=φ′(r)ψ′(r) + φ′′(r)ψ(r)

(φ′(r))2

and observe that since |φ′(r)| is bounded away from zero ψ1 is also a smooth function with

compact support in r ∼ 1. Apply the same argument to ψ1 as we did for ψ we getanother smooth function ψ2 with compact support in r ∼ 1 such that

ˆ ∞

0eiN

2tφ(r)ψ1(r)dr = − 1

iN2t

ˆ ∞

0eiN

2tφ(r)ψ2(r)dr.

By iterating this procedure k times we get a smooth function φk with compact support inr ∼ 1 such that

ˆ ∞

0eiN

2tφ(r)ψ(r)dr = (−1)k1

(iN2t)k

ˆ ∞

0eiN

2tφ(r)ψk(r)dr.

It follows that

|PnKt(x)| =∣∣∣∣∣N

d

ˆ

Sd

ˆ ∞

0eiN

2tφ(r)ψ(r)drdσ(ω)

∣∣∣∣∣

=

∣∣∣∣∣Nd

ˆ

Sd(−1)k

1

(iN2t)k

ˆ ∞

0eiN

2tφ(r)ψk(r)drdσ(ω)

∣∣∣∣∣

. Nd−2k

|t|k .

If d is even we take k = d2 and for d odd we take k = d+1

2 , and t N−2 by assumption sowe get the desired bound

|PnKt(x)| . 1

|t| d2.

DISPERSIVE PDE LECTURE 11 AND 12 3

Subcase 2b: |x| & tN . We use the following fact from Appendix b in Grafakos: For radialfunctions f(ξ) = f0(|ξ|), ξ ∈ Rd

f(x) =2π

|x| d−22

ˆ ∞

0f0(r)J d−2

2(2π|x|r)r d2 dr.

Note that f is also radial so in particular

f(x) = f(−x).

Now observe that applying the fact above to the function defined by f(ξ) = e−4π2it|ξ|2ψ(|ξ|N

)

and then a change of variables r 7→ Nr we get

PNKt(x) =

ˆ

Rde−2πix·(−ξ)e−4π2it|ξ|2ψ

( |ξ|N

)dξ

=2π

|x| d−22

ˆ ∞

0e−4π2it|ξ|2ψ

( |ξ|N

)J d−2

2(2π|x|r)r d2 dr

=2πN

d+22

|x| d−22

ˆ ∞

0e−4π2itN2|ξ|2ψ (|ξ|) J d−2

2(2πN |x|r)r d2 dr.

Using the Taylor expansion for Jv(s) we can write

Jv(s) =

√2

πscos(s− πv

2− π

4

)+O(s−

32 )

for large enough s. Recall that

cos(x) =eix + e−ix

2.

The integrand above is zero outside of a set which is bounded away from zero and |x| & tNby assumption, so we can choose N large enough to ensure that

J d−22

(2πN |x|r) =

√1

π2N |x|r

(e−i(

π(d−2)4

+π4

)e2πiN |x|r + ei(π(d−2)

4+π

4)e−2πiN |x|r

2

)+O((N |x|r)− 3

2 )

(1.2)in the region r ∼ 1.

First we deal with the error term O((N |x|r)− 32 ). Observe that

∣∣∣∣∣2πN

d+22

|x| d−22

ˆ

Rde−4π2itN2r2

ψ(r)O((N |x|r)− 32 )r

d2

∣∣∣∣∣ .N

d+12

|x| d+12

1

N. 1

|t| d+12

1

N. 1

|t| d2,

where the second inequality follows from the assumption that |x| & tN and the finalinequality from the assumption that t N−2.

Now we deal with the main contribution from (1.2). Consider the following integrals

I±t (x) =N

d+12

|x| d−12

ˆ ∞

0eitN

2φ(r)ψ(r)rd−1

2 dr,


where φ(r) = −4π2r2 ± 2π |x|tN r. Note that |φ′′(r)| = | − 8π2| ≥ 1, hence by a corollary ofthe Van der Corput Lemma we have

|I±t (x)| . Nd+1

2

|x| d−12

1

(|t|N2)12

. N

|t| d−12

1

|t| 12N=

1

|t| 12,

where the second inequality follows from the assumption that |x| & tN .It follows that for all dyadic N and t ∈ R

‖PNKt‖L∞x .1

|t| d2

2. Lecture 12

The goal of this lecture is to prove the following maximal function estimate which canbe used to prove a.e. convergence of the linear Schrodinger operator, in these notes we willonly prove a.e. convergence of S(t)f in C(Rt;Hs

x(Rd)) for d = 1 and s ≥ 14 .

Let Dα denote the fractional derivative operator of order α, more precisely for suitableregularity assumptions on f : Rd → R we define

Dαf(x) := F−1(

2π|ξ|αf(ξ)).

2.1. Maximal function estimate.

Proposition 2.1 (Maximal function estimate for d = 15). For d = 1 the following estimateholds,

‖S(t)f‖L4xL∞t. ‖D 1

4 f‖L2x.

There have been many partial results regarding the question of a.e. convergence for thelinear Schrodinger operator; when do we have

u(x, t) = S(t)f(x)→ f(x)

as t→ 0 for a.e. x ∈ Rd.The first result published in the 80’s is due to Carleson.

• Carleson ’80, a.e. convergence for f ∈ H 14 (R)

• Dahlberg-Kenig ’82, false if s < 14

• Sjolin-Vega ’85, a.e. convergence for s > 12 and ∀d

• Bourgain, Vargas-Vega, Tao-V-V, T-Vargas ’90’s, various partial results for partic-ular s < 1

2 and dimension d

• Sanghyuk Lee’06, a.e. convergence for s > 38 and d = 2

• Bourgain 12’,– d ≥ 3, sufficient condition for a.e. convergence is s > 1

2 − 14d

– d ≥ 4, necessary condition for a.e. convergence is s ≥ 12 − 1

2d• Luca-Rogers ’15, Bourgain• Du-Guth-Li 17’, a.e. convergence for s > 1

3 and d = 2 (sharp up to s = 13)

• D-G-L-Zhang ’18, a.e. convergence for d ≥ 3 and s > (d+1)2(d+2)

Now we will show how the maximal function estimate we stated above implies Carleson’soriginal result.


Proof of Proposition 2.1. We want to show that for d = 1 the following estimate holds,

‖S(t)f‖L4xL∞t. ‖D 1

4 f‖L2x

for f ∈ H14

(R). Let T = D−14S(t) which defines a linear operator T : L2

x → L4xL∞t . It

follows that the transpose T ∗ : L4xL∞t → L2

x is defined as follows,

T ∗G =

ˆ

RD−

14S(−t)G(t, x)dt.

Since ‖T‖ = ‖T ∗‖, it suffices to prove that

‖T ∗G‖L2x. ‖G‖

L43x L

1t

for G ∈ L43xL1

t . Furthermore we have TT ∗ : L4xL∞t → L4

xL∞t defined by

TT ∗G =

ˆ

RD−

12S(t− t′)G(t′, x)dt′.

Since ‖TT ∗‖ ≤ ‖T‖‖T ∗‖, it suffices to show that

‖TT ∗G‖L4xL∞t. ‖G‖

L43x L

1t

, (2.1)

for G ∈ L43xL1

t .Here we state a lemma which we will prove later,

Lemma 2.2. There exists some constant C > 0 independent of t such that∣∣∣∣ˆ

Rei(xξ−tξ

2) 1

|ξ| 12dξ

∣∣∣∣ ≤ C1

|x| 12.

Now we use this to establish (2.1). Recall that S(t)f = Kt ∗x f , where

Kt(x) =

ˆ

Re2πi(xξ−2πtξ2)dξ.

It follows that

‖TT ∗G‖L4xL∞t

=

∥∥∥∥ˆ

RD−

12K(·, x− y) ∗t G(·, y)dy

∥∥∥∥L4x

.

Young’s convolution inequality with respect to t and p =∞, q = 1, r =∞ tells us that∥∥∥∥ˆ

RD−

12K(·, x− y) ∗t G(·, y)dy

∥∥∥∥L4x

≤∥∥∥∥ˆ

R‖D− 1

2K(·, x− y)‖L∞t ‖G(·, y)‖L1tdy

∥∥∥∥L4x

.

Notice that

D−12K(·, x) =

1

2π

ˆ

Rei(xξ−tξ

2) 1

|ξ| 12dξ

So apply the above lemma we get∥∥∥∥ˆ

R‖D− 1

2K(·, x− y)‖L∞t ‖G(·, y)‖L1tdy

∥∥∥∥L4x

.∥∥∥∥∥

1

|x| 12∗x ‖G(t, x)‖L1

t

∥∥∥∥∥L4x

.

The Hardy-Littlewood-Sobolev inequality states that∥∥∥∥1

|x|d−α ∗ f∥∥∥∥Lq. ‖f‖Lp ,


provided that 1q = 1

p − αd . Applying the Hardy-Littlewood-Sobolev inequality with d =

1, p = 43 , q = 4, α = 1

2 gives∥∥∥∥∥

1

|x| 12∗x ‖G(t, x)‖L1

t

∥∥∥∥∥L4x

. ‖G‖L

43x L∞t

.

Thus we have shown that

‖TT ∗G‖L4xL∞t. ‖G‖

L43x L

1t

,

which completes the proof.

Now we return to proving Lemma 2.2, that is∣∣∣∣ˆ

Rei(xξ−tξ

2) 1

|ξ| 12dξ

∣∣∣∣ ≤ C1

|x| 12.

Firstly we consider two cases separately in order to establish the estimate.Case 1: |ξ| . |x|−1. Trivially we have

∣∣∣∣ˆ

|ξ|.|x|−1ei(xξ−tξ

2) 1

|ξ| 12dξ

∣∣∣∣ .ˆ c|x|−1

−c|x|−1

1

|ξ| 12dξ = 2

ˆ c|x|−1

0

1

ξ12

dξ = |ξ| 12∣∣∣∣c|x|−1

0

∼ 1

|x| 12.

Case 2: |ξ| |x|−1. Observe thatˆ

|ξ|&|x|−1ei(xξ−tξ

2) 1

|ξ| 12dξ =

ˆ

ξ&|x|−1ei(xξ−tξ

2) 1

|ξ| 12dξ +

ˆ

ξ.−|x|−1ei(xξ−tξ

2) 1

|ξ| 12dξ

ˆ

ξ&|x|−1ei(xξ−tξ

2) 1

|ξ| 12dξ +

ˆ

ξ&|x|−1ei(−xξ−tξ

2) 1

|ξ| 12dξ.

So it suffices to show that ∣∣∣∣ˆ

ξ&|x|−1

e−i(tξ2±xξ) 1

|ξ| 12dξ

∣∣∣∣ .1

|x| 12

By a change of variables ξ 7→ ξ12 we have

ˆ

ξ&|x|−1e−i(tξ

2±xξ) 1

|ξ| 12dξ =

ˆ

η&|x|−2e−i(tη±xη

12 ) 1

η34

dη.

Set Φ±(η) = −(txη ± η

12

). Now we consider two separate subcases.

Subcase 2a: |Φ±(η)| ≤∣∣ t

2x

∣∣. Observe that∣∣∣∣t

xη ± η 1

2

∣∣∣∣ ≤∣∣∣∣t

2x

∣∣∣∣ =⇒ η ∼(xt

)2

Note that |Φ′′±(η)| = η−32 and |Φ′′′±(η)| = η−

52 . Write the integral as

ˆ

η&|x|−2ei(xmin |Φ′′±(η)|)· Φ±

min |Φ′′±(η)| |Φ′′±(η)| 12dη.


Recall the Van Der Corput Lemma proved in an earlier lecture, applying the corollary we

stated with λ = xmin |Φ′′±(η)|, φ = Φ±min |Φ′′±(η)| and k = 2 we get

ˆ

η&|x|−2ei(xmin |Φ′′±(η)|)· Φ±

min |Φ′′±(η)| |Φ′′±(η)| 12dη . (xmin |Φ′′±(η)|)− 12

[max |Φ′′±|

12 +

ˆ

η∼(xt )

2 |Φ′′±|−

12 |Φ′′′±|dη

]

∼(x · t

3

x3

)− 12

[(t

x

) 32

+

(t

x

) 32(t

x

)5 (xt

)2]. 1

|x| 12.

Subcase 2b: |Φ±(η)| >∣∣ t

2x

∣∣. Observe that by the reverse triangle inequality we get

|Φ±(η)| ≥ η− 12 −

∣∣∣∣t

x

∣∣∣∣ > η−12 − 2|Φ±(η)|,

thus |Φ±(η)| . η−12 . Applying a similar integration by parts argument as we did earlier

yields ∣∣∣∣∣

ˆ

η&|x|−2eixΦ±(η) 1

η34

dη

∣∣∣∣∣ .1

|x| 12.

This completes the proof of the lemma.

2.2. Local smoothing estimate. Here we briefly introduce another estimate which com-bined with the maximal function estimate is used to study NLS with a derivative nonlin-earity.

Proposition 2.3 (Local smoothing). For d = 1 we have∥∥∥D 12S(t)f

∥∥∥L∞x L

2t

∼ ‖f‖L2x

Proof. By a change of variables η = ξ2 we can write

D12S(t)f(x) =

ˆ

|ξ| 12 ei(xξ−tξ2)f(ξ) dξ ∼ˆ

|η|− 14 ei(xη

12−tη)f(η) dη.

Applying Plancherel’s theorem first in t and the in x we get

∥∥∥D 12S(t)f

∥∥∥L2t

∼∥∥∥∥η−

14 eixη

12 f(η

12 )

∥∥∥∥L2η

=

(ˆ

1

|η| 12|f(η

12 )|2 dη

) 12

∼(ˆ

|f(ξ)|2 dξ) 1

2

= ‖f‖L2x.

Remark 2.4. • T.Tao gives a proof of local smoothing without using the Fourier

Transform in a short note entitled ”A physical space proof of the bilinear Strichartzand local smoothing estimates”.• For d ≥ 1, the local smoothing estimate becomes:

(ˆ

R

ˆ

Rd

∣∣∣∣|∇|12S(t)f(x

∣∣∣∣2

e−|x|2dx dt

) 12

. ‖f‖L2x.

See Oberwolfach lecture notes by Monica Vian (also ’16 NLS course notes by Hiro).• See ’16 NLS course notes by Hiro for LWP of fKdV

∂tu+ ∂3xu = ∂xu

5.

LECTURES 13 - 15

ANDREIA CHAPOUTO

1. Global-in-time behaviour of solutions to NLS

Using Strichartz estimates, it is possible to prove local well-posedness of NLS on Rd ifs ≥ max (scrit, 0), if p ∈ 2N + 1. Otherwise, an extra condition is required to prove theresult.

After guaranteeing LWP, it is reasonable to wonder about the global-in-time existenceof solutions. There are two possible answers:

• the solution exists globally in time (global well-posedness), or• it ceases to exist at some finite time (finite time blowup solution).

If u exists globally in time, then what is the behaviour of the solution u as t→ ±∞? Aninteresting phenomenon is that of scattering, when the solution approaches a linear solutionas t→ ±∞ t, in some sense ‘asymptotic linear behaviour’. To make the definition precise,

∃u± ∈ Hs(Rd)

s.t limt→±∞

‖u(t)− S(t)u±‖Hs = 0,

for some initial condition u±.Note that u± could be different from the initial condition u0 for the NLS and there could

be different asymptotic behaviour for t→ +∞ and t→ −∞.Reversely, one can focus on non-scattering solutions, such as solitons, solutions of the

form

u(t) = eitQ(x).

Note that

‖u(t)‖L∞x = ‖Q‖L∞x 6→ 0.

An open problem in the field is called the soliton resolution conjecture, i.e., for generic initialdata, u(t) decouples into a sum of solitons and radiation (scattering part) as t → ±∞. Ithas been shown for ‘integrable equations’ such as KdV and NLW (in the radial case, Kenig-Merle 2012).

Before we discussed the conservation of

Mass: M(u) =

ˆ

|u|2 dx,

Momentum: P (u) = Im

ˆ

u∇u dx,

Hamiltonian/Energy: H(u) =1

2

ˆ

|∇u|2 dx± 1

p+ 1

ˆ

|u|p+1 dx,

with ‘+’ corresponding to the defocusing case/repulsive case, and ‘−’ to the focusingcase/attractive case.

Note that the momentum is not as useful as the remaining conserved quantities, since itis vector-valued, therefore not sign definite.

1

2 ANDREIA CHAPOUTO

Example 1 (1-d cubic NLS, scrit = −1/2). Using LWP in L2(R) in the subcritical sense(T ∼ ‖u0‖−θL2

)and mass conservation, it is possible to prove GWP in L2(R).

However, this approach is not useful in R2, for which scrit = 0. T was given such that

‖S(t)u0‖L4TL

4x 1.

GWP in L2(R2)

holds true but the proof is much more complicated (Dodson 2012).

Example 2 (3-d defocusing cubic NLS, scrit = 1/2). LWP holds in H1(R3)

in the subcrit-ical sense (s = 1 > 1/2),

‖u(t)‖2H1 ≤ˆ

|u|2 dx+

ˆ

|∇u|2 dx+2

4

ˆ

|u|4 dx

= M (u(t)) + 2H (u(t))

= M (u0) + 2H (u0) <∞,using conservation of mass and energy. This result can be used to prove GWP in H1

(R3).

Note that the result is not true in the focusing case and that GWP (and scattering) in

H1/2(R3)

is an open problem.

Let us clarify some notation.

Definition 1.1. We say NLS is

• mass-critical if scrit = 0,• mass-subcritical if scrit < 0,• mass-supercritical if scrit > 0.

Similarly, we say NLS is

• energy-critical if scrit = 1,• energy-subcritical if scrit < 1,• energy-supercritical if scrit > 1.

For GWP and scattering of energy-critical defocusing NLS, check work by Bourgain(1999), CKSTT (1908) and Visan. For the energy-supercritical defocusing NLS, there is

LWP in Hscrit(Rd)

but GWP is an open problem even for smooth solutions.

1.1. Solitons and finite time blowup solutions.Consider the focusing NLS

i∂tu+ ∆u = −|u|p−1u.The NLS admits a special solution defined as follows.

Definition 1.2 (Solitary wave solution). We define u(t, x) = eitφ(x) as a soliton. It is asolution to NLS iff φ solves the following elliptic PDE

∆φ− φ+ |φ|p−1φ = 0, φ ∈ H1(Rd). (1.1)

Remark 1.3. Note that for d = 1, all solutions to (1.1) are translates of

Q(x) =

p+ 1

2 cosh2((p−1)x

2

)

p−1

.

For d ≥ 2, there exists a sequence Qnn≥0 of real radial solutions to (1.1) with increasing

L2-norms such that Qn(r) vanishes n times on R+.

LECTURES 13 - 15 3

There exists Q0 radially symmetric and positive, called the ground state (Berestycki-Lions-Peletier 1981). In addition, there exists a unique φ > 0, φ ∈ H1, radial and C2 withexponential decay (Gidas-Ni-Nirenberg 1979, Kwong 1989). To determine the sequence onecan use the shooting method in r > 0.

Lastly, ground states play an important role in elliptic, parabolic and dispersive PDEs,variational problems and functional inequalities.

According to Definition 1.1, NLS is mass-subcritical if p < 1 + 4d , scrit = 0, and we define

the NLS scaling applied to Q

Qλ(x) := λ2

p−1Q(λx).

Proposition 1.4 (Variational characterization of Q). Let d ≥ 1, 1 < p < 1 + 4d , M > 0

fixed. Then, the minimization problem

min‖u‖L2=M

H(u)

has minimum attained atQλ(M) (· − x0) eiγ0 ,

for all x0 ∈ Rd, γ0 ∈ R, and with λ(M) ∈ R chosen such that∥∥Qλ(M)

∥∥L2 = M .

In order to prove this proposition, one can focus on the Gateaux derivative of H,

d

dεH(u+ εv)

∣∣∣∣ε=0

.

Recall that for the mass-critical case scrit = 0 and p = 1 + 4d . Let

J(u) =

(´

|∇u|2) (´

|u|2)2/d

´

|u|2+4/d,

for u 6= 0.

Proposition 1.5.

(1) The minimization problemminu∈H1

u6=0

J(u)

has minimum attained at

λd/20 Q (λ0x+ x0) e

iγ0 ,

for any (λ0, x0, γ0) ∈ R+ × Rd × R. In particular, we have the following sharpGagliardo-Nirenberg inequality:

ˆ

|u|2+4/d ≤ (J(Q))−1︸︷︷︸optimal constant

ˆ

|∇u|2(ˆ

|u|2)2/d

.

(2) (Rigidity) Let u ∈ H1 such thatˆ

|u|2 =

ˆ

Q2,

H(u) = 0(

= H(Q)).

Then, u(x) = λd/20 Q (λ0x+ x0) e

iγ0 for some (λ0, x0, γ0) ∈ R+ × Rd × R.

4 ANDREIA CHAPOUTO

Remark 1.6. The sharp Gagliardo-Nirenberg inequality falls directly from the fact thatJ(Q) ≤ J(u), for all u ∈ H1.

Note that for all u ∈ H1,

H(u) ≥ 1

2

ˆ

|∇u|2(

1−( ‖u‖L2

‖Q‖L2

)4/d)dx.

Let u0 ∈ H1 such that ‖u0‖L2 < ‖Q‖L2. By mass conservation,

H(u) ≥ c1ˆ

|∇u|2 dx =⇒ H(u) +M(u) ≥ c2‖u‖2H1 ,

with c1, c2 > 0 constants independent of t, but which depend on ‖u0‖L2.

Using this result, it is possible to prove GWP in H1(Rd), provided that ‖u0‖L2 < ‖Q‖L2,

as well as scattering. However, mass-subcritical NLS is GWP in H1(Rd) regardless of‖u0‖L2.

Now we want to consider what happens when the initial condition has the mass of thesoliton Q, i.e., ‖u0‖L2 = ‖Q‖L2 . The soliton u(t) = eitQ is a global non-scattering solution.

When scrit = 0 we can apply the pseudo-conformal symmetry to u,

u(t, x) 7−→ v(t, x) =1

|t|d/2u(−1

t,x

t

)ei|x|24t , (t 6= 0).

Applying the pseudo-conformal symmetry to eitQ we obtain

Q∗(t, x) =1

|t|d/2Q(xt

)e−i

|x|24t

+ it ,

which is a solution to NLS for t < 0. Considering Q∗ starting at some negative time, itblows up at time t = 0. In some sense

|Q∗(t)|2“→ ”‖Q‖2L2δx=0 as t 0

‖∇Q∗(t)‖L2 ∼ 1

|t| (blowup speed).

Therefore, Q∗ is the minimal mass blowup solution, which is unique (Merle 1993), because ifu is such that M(u) = M(Q) and u blows up in finite time, then u = Q∗ up to symmetries.Using scaling, it is also possible to determine a lower bound for the blowup rate, & 1√

T ∗−t .If M(Q) < M (u0) < M(Q) + ε, there exists another finite time blowup solution called

“log log” blowup solution, which is stable and has blowup rate ∼√

log log(T ∗−t)T ∗−t (Merle-

Raphael 2000’s).

1.2. Virial identity & Morawetz estimate. Once again, consider NLS

i∂tu+∇u = λ|u|p−1u, λ = ±1,

with λ = 1 defocusing, and λ = −1 focusing case. We can rewrite the equation to have

∂tu = i∆u− iλ|u|p−1u.Definition 1.7. We define the virial potential as

Va(t) =

ˆ

a(x)|u|2 dx,

for some function a.

LECTURES 13 - 15 5

We start by computing the first and second derivatives of Va(t),

∂tVa(t) =

ˆ

a(x)∂t|u|2 dx

=

ˆ

a(x)2 Re (∂tu · u) dx

= 2

ˆ

a(x) Re((i∆u− iλ|u|p−1u

)u)dx

= 2

ˆ

a(x) Re (i∆u u) dx,

∂2t Va(t) = 4

ˆ

Re (∂ku∂j u) ∂k∂ja+ 2λp− 1

p+ 1

ˆ

|u|p+1∆a−ˆ

|u|2∆2a,

using Einstein’s summation notation for the first integral, i.e., summing over repeatedindices.

The following examples present two different choices for function a, which lead to thevirial identity and Morawetz estimate, respectively.

Example 3 (Virial identity). Let a(x) = |x|2 =d∑j=1

x2j . Then,

∆a = 2d,

∆2a = 0,

∂k∂ja = 2δjk,

with δjk the Kronecker delta. We call

V (t) =

ˆ

|x|2|u(t, x)|2 dx

the virial identity, which can be seen as a variance, with |u(t)|2 dx as a measure. Thus,

∂2t V (t) = 8

ˆ

|∇u|2 + 4dλp− 1

p+ 1

ˆ

|u|p+1

= 16H(u) +4λd

p+ 1

(p−

(1 +

4

d

)

︸︷︷︸mass-critical

power

)ˆ|u|p+1.

Therefore, in the mass-critical case,

∂2t V (t) = 16H(u),

which is conserved. If H (u0) < 0, then ∂2t V (t) = 16H (u0) < 0, showing that V (t) is adownward parabola. Thus, V (t∗) < 0 for some t∗ > 0, but

V (t) =

ˆ

|x|2|u(t)|2 dx ≥ 0,

reaching a contradiction. Then, u must blow up before time t∗ (Glassey’s argument, Za-kharov’s).

A similar argument holds for the mass-supercritical case.

6 ANDREIA CHAPOUTO

Example 4 (Morawetz estimate). Let a(x) = |x|. Then, ∂ja =xj|x| and

∂2j a =1

|x| −xj|x|2

xj|x| =⇒ ∆a(x) =

d− 1

|x| .

Similarly,

∆2a =

− (d−1)(d−3)

|x|3 ≤ 0 if d > 3

−8πδ if d = 3

Combining these results with the expression for ∂2t Va(t) gives

∂2t Va(t) = 4

ˆ |∇u|2|x| −

∣∣∣ x|x| · ∇u∣∣∣2

|x| dx+ 2λp− 1

p+ 1

ˆ |u|p+1

|x| dx−ˆ

|u|2∆2a dx.

On the other hand,

∂2t Va(t) = ∂t2 Im

ˆ

∇u · x|x| u.

For the defocusing case, λ = 1,ˆ t2

t1

ˆ |u|p+1

|x| dx dt . supt=t1,t2

∣∣∣∣Imˆ

∇u · x|x| u dx∣∣∣∣

. supt1,t2‖u(t)‖2

H1/2

or

.M (u0)1/2H (u0)

1/2 .

Then, one can send t1 → −∞ and t2 → +∞ to obtain Morawetz estimate for the defocusingcase

ˆ

Rt

ˆ

Rdx

|u(t, x)|p+1

|x| dx dt . supt‖u(t)‖2

H1/2

or

.M (u0)1/2H (u0)

1/2 .

By repeating the derivation centered at y, for d = 3, we get

∂t Im

ˆ

∇u(x) · x− y|x− y| u(x) dx = 2

ˆ |∇u(x)|2|x− y| −

∣∣∣ x−y|x−y| · ∇u(x)∣∣∣2

|x− y| dx

+ 2λp− 1

p+ 1

ˆ |u(x)|p+1

|x− y| dx

+ 4π|u(y)|2.

(1.2)

Multiplying (1.2) by |u(y)|2 and integrating with respect to y, gives the Interaction Morawetzestimate (Colliander-Keel-Staffilani-Takaoka-Tao, 2000)

ˆ

Rt

ˆ

R3x

|u(y)|4 dy dt . supt‖u(t)‖2L2‖u(t)‖2

H1/2

or

.M (u0)3/2H (u0) .

LECTURES 13 - 15 7

1.3. Scattering for energy-subcritical defocusing cubic NLS on R3 (GWP inH1(R3)). Let us focus on the case t → +∞. We want to show that there exists u+ ∈

H1(R3)

such that

‖u(t)− S(t)u+‖H1 → 0 as t→ +∞,which is equivalent to the statement with respect to norm ‖S(−t)u(t)− u+‖H1 , with theoperator S(−t) as follow

S(−t)u(t) = u0 − iˆ t

0S(−t′)|u|2u(t′) dt′.

It suffices to make sense of∞

0

S(−t)|u|2u(t) dt, in H1.

We start by focusing on the existence of the wave opetaor Ω+ : u+ ∈ H1 → u0 ∈ H1,i.e., given u+ ∈ H1, we want to know if we can find u0 ∈ H1 such that the correspodingsolution u scatters to S(t)u+.

Remark 1.8. If Ω+ exists, it is injective, as a consequence of the uniqueness part of thewell-posedness theory.

If Ω+ is invertible, we say that we have asymptotic completeness.

Let u+ and S(−t)u(t) as follows,

u+ = u0 − i∞

0

S(−t′)|u|2u(t′) dt′,

S(−t)u(t) = u0 − it

ˆ

0

S(−t′)|u|2u(t′) dt′.

Applying S(t) to both and taking a difference

u(t) = S(t)u+ + i

∞

t

S(t− t′)|u|2u(t′) dt′, (1.3)

leads to a terminal value problem.

Theorem 1.9 (Existence of wave operator). The wave operator Ω+ exists in S′ ([T,∞)).

Proof. Well-posedness on [T,∞) Let S′ = L5t,x∩L10/3

t W1,10/3x , with (10/3, 10/3) admissible.

Then, using Sobolev in space

‖u‖L5t,x. ‖u‖

L5tW

1,30/11x

,

with (5, 30/11) also admissible. By Strichartz estimates,

‖S(t)u+‖S′(Rt). ‖u+‖H1 <∞

(11

30− 1

5=

1

6≤ 1

3

).

By MCT, there exists T > 0 such that

‖S(t)u+‖S′([T,∞)) ≤ ε.

8 ANDREIA CHAPOUTO

Define Γu(t) as the right-hand side of (1.3), then considering u, v ∈ B2ε ⊂ S′ ([T,∞)),∥∥∥Γu

∥∥∥S′([T,∞))

≤ ε+ C∥∥〈∇〉

(|u|2u

)∥∥L10/7t,x ([T,∞))

. ε+ ‖u‖2L5t,x‖〈∇〉u‖

L10/3t,x

≤ ε+ ‖u‖3S′([T,∞))

≤ ε+ C (2ε)3 . (1.4)

Similarly,∥∥∥Γu− Γv

∥∥∥S′([T,∞))

≤ C(‖u‖2

S′([T,∞))+ ‖v‖2

S′([T,∞))

)‖u− v‖S′([T,∞)). (1.5)

Consequently, we can choose ε 1 such that (1.4) and (1.5) prove that Γ is a contraction

on B2ε ⊂ S′ ([T,∞)). Now, we can apply local well-posedness and the conservation of massand Hamiltonian to extend u onto [0, T ].

In summary, for [T,∞) one uses small data theory, and for [0, T ] iteration of the LWPtheory.

We now focus on proving scattering (asymptotic completeness).

∥∥∥∥ˆ ∞

0S(−t)|u|2u(t) dt

∥∥∥∥H1

.∥∥〈∇〉

(|u|2u

)∥∥L10/7t,x

(dual Strichartz)

. ‖u‖2L5t,x‖u‖

L10/3t W

1,10/3x

≤(

sup(q,r) admissible

‖〈∇〉u‖LqtL

rx

)3

= ‖u‖3S1 .

Then, the strong space-time bound ‖u‖S′(Rt) <∞ implies scattering.

Claim 1.10. There exists a “weak” space-time bound

‖u‖Lqt,x. 1,

for some q ∈[103 , 10

], which implies “strong” space-time bound ‖u‖S1 . 1 (which in turn

implies scattering).

Remark 1.11. Note that interaction Morawetz implies “weak” space-time bound (q = 4),while Morawetz estimate in the radial setting implies ‖u‖L5

t,x. 1.

Using radial Sobolev inequality

‖|x|s|u|‖L∞x (Rd) . ‖u‖H1 ,

for d2 − 1 ≤ s ≤ d−1

2 ,ˆ

t

ˆ

x|u|5 dx dt =

ˆ

t

ˆ

x|x||u| · |u|

4

|x| dx dt ≤ ‖|x||u|‖L∞t,x︸︷︷︸

radial Sobolev

ˆ

t

ˆ

x

|u|4|x|︸︷︷︸

Morawetz

≤ C (‖u0‖H1) .

Proof of Claim 1.10. Given ε > 0, divide R+ = ∪Nj=1Ij , with N ∈ Z≥0, such that

‖u‖LqIjLqx≤ ε.

LECTURES 13 - 15 9

Note that this implies‖u‖Lq

tLqx(Rt×Rd

x)<∞.

With Ij = [tj , tj+1),

u(t) = S(t− tj)u(tj)− iˆ t

tj

S(t− t′)|u|2u(t′) dt′, t ∈ Ij ,

which implies, using Strichartz,

‖u(t)‖S1(Ij) . ‖u(tj)‖H1 + ‖u‖2L5IjL5x‖u‖

L103

IjW

1, 103x

≤ ‖u(tj)‖H1 + ‖u‖2L5IjL5x‖u‖S1(Ij). (1.6)

Note that

‖u‖L

103

t,x

≤ ‖u‖S0(I) ((10/3, 10/3) admissible)

≤ ‖u‖S1(I),

‖u‖L10t,x. ‖u‖

L10I W

1, 3013x

(using Sobolev)

≤ ‖u‖S1(I).

(13

30− 1

10=

1

3=s

d

)

By interpolation,

‖u‖L5t,x(Ij)

≤ ‖u‖θLqt,x(Ij)

‖u‖1−θL10t,x(Ij)

or

≤ ‖u‖θLqt,x(Ij)

‖u‖1−θL

103

t,x(Ij),

which implies‖u‖L5

t,x(Ij)≤ εθ‖u‖1−θ

S1(I). (1.7)

From (1.6) and (1.7), we get

‖u‖S1(Ij) . ‖u(tj)‖H1 + ε2θ‖u‖3−2θS1(Ij)

.

Now, let X(t) = ‖u‖S1([tj ,t]), for t ∈ Ij . We want to prove

X(t) . 1 + ε2θX(t)3−2θ, t ∈ Ij .Note that X(t) is continuous in t but it does not approach 0 as t → tj , and X(tj) . 1.

Intuitively, considering the curve y = c + cε2θx3−2θ, for 0 < θ < 1 and 0 < ε 1, weare concerned with the values of x such that the first curve lies above y = x. These willcorrespond to two portions of the positive axis, one of those close to the origin. As aconsequence of the continuity of X(t) it is not possible to move from this set to the one farfrom the origin. Therefore, X(t) . 1 and

X(t) . 1 + ε2θX(t)3−2θ, ∀t ∈ Ij .Thus, ‖u‖S1(Ij) . 1 and summing over finitely many intervals Ij , we obtain

‖u‖S1([0,∞]) . 1,

which implies scattering.

DISPERSIVE EQUATIONS: LECTURE NOTES

STEFANIA LISAI

Abstract. We explore different scenarios that arise in case of ill-posedness of NLS in lowregularity and show some examples. These notes cover Lecture 16 and half of Lecture 17of the MIGSAA Advanced course Dispersive Equations, taught in 2018.

1. Ill-posedness of NLS in low regularity

In this section, we focus on the concept of ill-posedness for the nonlinear Schrodingerequation (NLS)

i∂tu+ ∆u± |u|p−1u = 0

u|t=0 = u0 ∈ Hs(M)(1.1)

where M = Rd or M = Td. We list here some possible scenarios of bad behaviour of NLSin low regularity (s ≤ sc or s < sc):

(1) Failure of the nonlinear estimate.Even if the linear estimate∥∥∥∥

ˆ t

0S(t− t′)|u|p−1u(t′) dt′

∥∥∥∥XsT

.∥∥|u|p−1

∥∥NsT,

holds, the nonlinear estimate∥∥|u|p−1u∥∥NsT. ‖u‖pXs

T

might fail for s < sc. We used here the notation XsT ⊂ CTHs

x for the solution space.

Example 1 (2-d cubic NLS). The scaling critical regularity for the cubic (p = 3)NLS on R2 is sc = d

2 − 2p−1 = 0, and the nonlinear estimate

∥∥〈∇〉s(|u|2u)∥∥L4/3T,x

. ‖〈∇〉su‖3L4T,x

fails for s < 0.

(2) Failure of Ck-smoothness of the solution map

Φ : u0 ∈ Hs 7→ u ∈ CTHs.

Example 2. If p ∈ 2N+ 1, the nonlinearity |u|p−1u is algebraic and Φ is algebraicif we can solve the fixed point problem Γu0(u) = u by the standard contractionargument.

Remark 1.1. Failure of Ck-smoothness does not imply ill-posedness, but it saysthat the contraction argument cannot be used to show well-posedness. For instance,one would need to use more robust energy methods instead (e.g. in the context ofthe short-time Fourier restriction norm method).

(3) Failure of the uniform continuity of the solution map on bounded sets in Hs (i.e.failure of local uniform continuity). Like the condition (2), this does not implyill-posedness. This condition is called mild ill-posedness.

1

2 S. LISAI

(4) Failure of continuity of the solution map. This implies ill-posedness, by definitionof well-posedness.

(5) Failure of existence and/or uniqueness.

1.1. Failure of C3-smoothness for cubic NLS. We focus on the case of cubic NLS onHs(Td) for s < 0. We consider the cubic NLS with initial datum in Hs(Td):

i∂tu+ ∆u± |u|2u = 0

u|t=0 = δφ

for some smooth φ ∈ Hs(Td) and δ ∈ R. We consider the solution u(t, x; δ) depending onthe parameter δ. Observe that u(·, ·; 0) ≡ 0.Let Φ(t) : u0 ∈ Hs 7→ u(t) ∈ Hs be the solution map and observe that it is well-defined onsmooth functions. We want to prove that Φ /∈ C3: suppose Φ ∈ Ck([0, τ)) for small τ > 0and for some k ∈ N. We use the notation u0(δ) = u(0, ·; δ). By the smoothness around thezero function u0(0), we have that

Φ(t)(δφ) = Φ(t)(u0(δ)) = Φ(t)(0) +∇Φ(t)(0) · δφ+1

2∇2Φ(0)(δφ, δφ) + ... .

Using the chain rule

∂kδΦ(t)(δφ) = ∇kΦ(t)(δφ) · (φ, . . . , φ)

we have that∥∥∥∥dk

dδkΦ(δφ)

∣∣∣∣δ=0

∥∥∥∥Hs

.∥∥∥∇kΦ(t)(0)

∥∥∥Hs×k→Hs︸︷︷︸

≤C<∞

‖φ‖kHs .

Namely,∥∥∥∂kδΦ(t)(δφ)

∣∣δ=0

∥∥∥Hs

. ‖φ‖kHs . (1.2)

By Duhamel’s formula, we have that

u(t) = δS(t)φ± iˆ t

0S(t− t′)|u|2u(t′) dt′.

As the second term depends on δ with order > 1, we have that

∂u

∂δ

∣∣∣∣δ=0

= S(t)φ

∂2u

∂δ2

∣∣∣∣δ=0

= ±ˆ t

0S(t− t′)∂2δ

(|u|2u(t′)

) ∣∣δ=0

dt′ = 0,

where the last equality is given by the fact that applying the chain rule we obtain a sumof terms that contain at least one u|δ=0 = 0 or one u|δ−0 = 0. The third derivative in δ isnon-zero: the terms that include either u|δ=0 or u|δ−0 are zero, so we are left with

∂3u

∂δ3

∣∣∣∣δ=0

∼ˆ t

0S(t− t′)u2δ uδ

∣∣δ=0

dt′ +ˆ t

0S(t− t′)uδu2δ

∣∣δ=0

dt′

=

ˆ t

0S(t− t′)|S(t′)φ|2S(t′)φdt′ +

ˆ t

0S(t− t′)|S(t′)φ|2S(t′)φdt′.

DISPERSIVE EQUATIONS 3

Given N ∈ N, let φ(x) = N−seiNx1 , therefore ‖φ‖Hs ∼ 1 for any N ∈ N.We can write explicitly

S(t)φ = N−seiNx1−iN2t

=⇒ |S(t′)φ|2S(t′)φ = N−3seiNx1−iN2t′

=⇒ S(t− t′)φ(|S(t′)φ|2S(t′)φ

)= N−3seiNx1−iN

2t︸︷︷︸

indep. on t′

,

and similarly for the other term. Therefore the third derivative ∂3δu(0) is given by

∂3u

∂δ3

∣∣∣∣δ=0

∼ tN−3seiNx1−iN2t

and its norm is ∥∥∥∥∂3u

∂δ3

∣∣∣∣δ=0

∥∥∥∥Hs

∼ tN−3s.

Since ‖φ‖ ∼ 1, the estimate (1.2) cannot hold for s < 0, therefore Φ /∈ C3.This argument first appeared in KdV and mKdV context, in [1].

1.2. Failure of uniform continuity. We show now that we can construct a family ofpairs of smooth initial data (u0,ε, v0,ε)ε>0 (and their solutions (uε, vε)ε>0) such that forgiven ε > 0 and t > 0 there exist u0,ε and v0,ε such that

‖u0,ε − v0,ε‖ < ε but ‖uε − vε‖ & 1.

This technique has been used both in the case of whole space and in periodic setting:Kenigand, Ponce and Vega used a focusing family of soliton solutions on R with parametersin [5], Christ, Colliander and Tao used a defocusing family of approximate solutions on Rin [3], and Burq, Gerard and Tzvetkov worked on d-dimensional cubic NLS on periodicspace in [2]. In [2], the authors consider the special case d = 1 first, and deduce the resultfor general d considering u(x) = u(x1, 0, . . . , 0).

We consider hereuN,a(t, x) := aei(Nx−N

2t±|a|2t),for given N ∈ N and a ∈ C. For each a ∈ C and N ∈ N, uN,a is a solution of the 1−D cubicNLS. Choose a := N−sα and a′ := N−sα′, with α, α′ ∈ C, then

∥∥uN,a(0)− uN,a′(0)∥∥Hs ∼ |α− α′|.

Moreover, for a small t0 > 0 we have that

‖uN,a(t0)− uN,a′(t0)‖Hs ∼∣∣∣αe±iN−2s|α|2t0 − α′e±iN−2s|α′|2t0

∣∣∣

∼∣∣∣α− α′e±iN−2s(|α|2−|α′|2)t0

∣∣∣ .

We can choose N 1 such that ±iN−2s(|α|2 − |α′|2)t0 ∼ iπ + o(1), so

‖uN,a(t0)− uN,a′(t0)‖Hs ∼ |α|+ |α′|.We choose α and α′ such that

|α|, |α′| ∼ 1 and |α− α′| < ε.

Note that

|α|2 − |α′|2 =(|α| − |α′|

) (|α|+ |α′|

)∼ ε

4 S. LISAI

therefore we must have that N ∼ ε− 12s →∞ as ε→ 0.

Hence,

‖uN,a(t)− uN,a′(t)‖Hs

< ε if t = 0

∼ 1 if t = t0 > 0,

and the solution map is not uniformly continuous.

1.2.1. The importance of s = 0 for NLS. We showed “mild ill-posedness“ of cubic NLS inHs(Td) for s < 0. Even though the scaling critical regularity for cubic NLS is

scrit =d

2− 2

p− 1=d

2− 1,

and hence scrit = 0 if and only if d = 2, we have a threshold at s = 0 in any dimension, notjust for d = 2. In order to see what the role of s = 0 is, we look at Galilean invariance:given u a solution of NLS on Rd, then

uβ(t, x) := eiβ2·xe−

|β|24tu(t, x+ βt)

is also a solution, for β ∈ 2Zd. On the Fourier space

uβ(t, ξ) = e−i34|β|2teiξ·βtu(t, ξ − β

2).

Therefore

|uβ(t, ξ)| = |u(t, ξ − β

2)|

and the L2x-norm is invariant under Galilean symmetry, i.e. s∞crit = 0 is another critical

regularity for NLS associated with Galilean symmetry. Moreover, the Fourier-Lebesguespace FLs,p, where

‖f‖FLs,p = ‖〈ξ〉sf(ξ)‖Lpξ ,is invariant under Galilean symmetry when s = 0.

1.2.2. The 1-dimensional case. In the 1-d case, the critical regularity is scrit = −12 < 0, so

we would hope to have no problems above the threshold s = sc = −12 . Instead, for s < 0 we

have failure of the uniform continuity of the solution operator (”mild ill-posedness“) evenabove the critical regularity scrit = −1

2 , whereas the problem of local well-posedness is stillopen on R. Only existence has been proved on R, see [4], [6] and [7].On the 1-dimensional torus T, the cubic NLS is proved to be ill-posed for s < 0:

• in [3] discontinuity of the solution map in Hs(T) is proved for s < 0.• in [9] discontinuity of the solution map is proved from L2

x(T) endowed with the weaktopology into D′(T). This implies discontinuity on Hs(T) for s < 0.• In [8] non-existence of solutions is proved for initial datum u0 /∈ L2(T).

On the torus, the appropriate equation to study outside L2(T) is the Wick ordered NLS(WNLS) or renormalised NLS :

i∂tu+ ∂2xu±(|u|2 −

|u|2 dx)u = 0.

The integral term does not make sense if u /∈ L2(T), but if u ∈ C(R;L2(T)) solves NLS,then

G(u) := e2itffl

|u|2 dx

DISPERSIVE EQUATIONS 5

solves WNLS. The operator G is invertible on C(R;L2(T)), but it is not defined for initialdata outside of L2(T). The WNLS is easier to study outside L2(T) than NLS. An apriori estimates on its solution and invertibility of G can be used to prove non-existence ofsolutions to NLS in Hs(T) for s < 0.When WNLS is studied outside L2(T), it is written in the following form

i∂tu+ ∂2xu± (|u2| − 2 · ∞)u = 0.

References

[1] J. Bourgain, On the compactness of the support of solutions of dispersive equations, InternationalMathematics Research Notices, 9 (1997), 437–447.

[2] N. Burq, P. Gerard, N. Tzvetkov, An instability property of the nonlinear Schrodinger equation on Sd,Mathematical Research Letters, 9 (2002), no. 2-3, 323–335.

[3] M. Christ, J. Colliander, T. Tao, Asymptotics, frequency modulation, and low regularity ill-posednessfor canonical defocusing equations, American Journal of Mathematics, 125 (2003), no. 6, 1235–1293.

[4] M. Christ, J. Colliander, T. Tao, A priori bounds and weak solutions for the nonlinear Schrodingerequation in Sobolev spaces of negative order, Journal of Functional Analysis, 254 (2008), no. 2, 368–395.

[5] C. Kenigand, G. Ponce, L. Vega, On the ill-posedness of some canonical dispersive equations, DukeMathematical Journal, 106 (2001), no. 3, 617–633.

[6] H. Kochand and D. Tataru, A priori bounds for the 1D cubic NLS in negative Sobolev spaces, Interna-tional Mathematics Research Notices. IMRN (2007), no. 16.

[7] H. Kochand and D. Tataru, Energy and local energy bounds for the 1-d cubic NLS equation in H−14 ,

Annales de l’Institut Henri Poincare. Analyse Non Lineaire, 29 (2012), no. 6, 955–988.[8] Z. Guo, T. Oh, Non-Existence of Solutions for the Periodic Cubic NLS below L2, International Math-

ematics Research Notices. IMRN (2018), no. 6, 1656–1729.[9] L. Molinet,On ill-posedness for the one-dimensional periodic cubic Schrodinger equation, Mathematical

Research Letters, 16 (2009), no. 1, 111–120

LECTURES 17.6-19.5

WILLIAM J. TRENBERTH

Abstract. In this section of the notes we outline the result in [2]. That is for, s ≤ − 12

for d = 1 and s < 0 for d ≥ 2, we prove norm inflation for NLS for general initial data inHs(M) whereM = Rd or Td. As a result we show that under these regularity conditionsNLS is ill-posed.

1. Norm Inflation for the Cubic Nonlinear Schrodinger equation

1.1. Introduction and Statement of Results. In this section of the notes we will bestudying the cubic nonlinear Schrodinger equation (NLS)

i∂tu = ∆u± |u|2uu(0) = u0 ∈ Hs.

(1.1)

In previous sections we studied the local well-posedness of (1.1). In particular we provedthat (1.1) is locally well-posed for initial data in L2(M) where M = R,T. Here we meanwell-posedness in the Hadamard sense. Broadly this means:

• The solution exists locally in time (in some sense).• The solution is unique (in some space).• The solution map is continuous (with some topology).

In this section of the notes we will prove that (1.1) is ill-posed under certain conditions.We will prove ill-posedness by showing the solution map for (1.1) is not continuous underthese conditions.We will show the solution map is not continuous by proving ‘Norm Inflation’. The firstresult of this type was proven by Christ-Colliander-Tao.

Theorem 1.1. Norm inflation [1]: Given ε > 0 there exists a solution uε to (1.1) andtε ∈ (0, ε) such that

‖uε(0)‖Hs < ε but ‖uε(tε)‖Hs >1

ε. (1.2)

This result shows the solution map is not continuous at 0. In [2] this result was generalizedto general initial data.

Theorem 1.2. Norm inflation at general initial data [2]: Let d ≥ 1 and M = Rd or Td.Suppose that s ∈ R satisfies

(1) s ≤ −12 = scrit when d = 1

(2) s < 0 when d ≥ 2 (when d = 2, scrit = 0).

Fix u0 ∈ Hs(M). Then given ε > 0 there exists a solution uε to (1.1) and tε ∈ (0, ε) suchthat

‖uε(0)− u0‖Hs < ε but ‖uε(tε)‖Hs >1

ε. (1.3)

This shows that the solution map is not continuous everywhere!1

2 W. J. TRENBERTH

1.2. Power Series Expansion Indexed by Trees. As is usual we work with theDuhamel formulation of (1.1):

u(t) = S(t)u0 ± iˆ t

0S(t− t′)|u|2u(t′) dt′ = S(t)u0 + I[u, u, u] (1.4)

where

I[u1, u2, u3] = ±iˆ t

0S(t− t′)u1u2u3(t′) dt′. (1.5)

For convenience we use the notation I[u, u, u] = I[u]. In previous sections we proved LWPin Hs(M) for M = Rd,Td and s > d

2 using the algebra property for Hs(M). A similar

argument yields local well-posedness in the Weiner algebra FL1(M) = FL0,1(M) withtime of existence T ∼ ‖ · ‖−2

FL0,1(M). Here FLs,p(M) refers to the Fourier Lebesgue spaces

with norm

‖f‖FLs,p = ‖〈ξ〉sf(ξ)‖Lpξ . (1.6)

This implies that the Picard iterations converge. That Pj(φ) converges where,

P0(φ) = S(t)φ, Pj(φ) = S(t)φ+ I[Pj−1(φ)]. (1.7)

We define a Ternary tree to be a tree where each node either has no children (is a terminalnode) or has exactly three children (is a non-terminal node).

Define T (j) to be the collection of trees of the jth generation. t(J) is the collection oftrees of size T = 3j+1. Fix φ ∈ FL1. Given T ∈ T (j) where j ≥ 0, associate a multilinearoperator (in φ) by

• replace a non-terminal node by the Duhamel integral operator I[u1, u2, u3] whereu1, u2, u3 are the three childern.• replace a terminal node by the linear solution S(t)φ.

Denote this map by Ψ :∞⋃j=0

T (j) 7→ D′([−T, T ]×M).

Formally we have the following power series expansion for the solution, u, of (1.1).

u =∞∑

j=0

Ξj(φ) =∞∑

j=0

∑

T ∈T (j)

Ψφ(T ). (1.8)

Lemma 1.3. Let T (j) be as defined previously. Then |T (j)| ≤ cj0 for som ec0 > 0.

Proof. We proceed by induction. Note that |T (0)| = |T (1)| = 1. Let j ≥ 2 Then

|T (j)| =∑

j1+j2+j3=j−1j3,j2,j1≥0

|T (j1)||T (j2)||T (j3)|. (1.9)

By symmetry,

|T (j)| = 6∑

j1+j2+j3=j−1j3≥j2≥j1≥0

|T (j1)||T (j2)||T (j3)|. (1.10)

Assume a stronger bound

|T (j)| ≤ cj0(1 + j)2

, for all k ≤ j − 1. (1.11)

LECTURES 17.6-19.5 3

Then, using the fact that 3j3 = 3 max(ji + 1) ≥ j + 1,

|T (j)| = 6∑

j1+j2+j3=j−1j3,j2,j1≥0

3∏

i=1

Cji0

(1 + ji)2

≤ 6 · 32

∑

j1+j2+j3=j−1j3,j2,j1≥0

1

(1 + j1)2

1

(1 + j2)2

cj−10

(j + 1)2.

The result follows by setting

c0 = 6 · 32

∑

j2,j1≥0

1

(1 + j1)2

1

(1 + j2)2

. (1.12)

1.3. Basic Multilinear Estimates.

Lemma 1.4. Let j ≥ 0 and t ≥ 0. Then,

‖Ξj(φ)(t)‖FL1 ≤ Cjtj‖φ‖2j+1FL1

and‖Ξj(φ)(t)‖FL∞ ≤ Cjtj‖φ‖2j−1

FL1 ‖φ‖2L2 .

Lemma 1.5. Let j ≥ 0 and t ≥ 0. Then,

‖Ξj(u0 + φ)(t)− Ξj(φ)(t)‖FLp ≤ Cjtj‖u0‖FLp(‖u0‖2jFL1 + ‖φ‖2jFL1

)(1.13)

We will now begin the proof of Theorem 1.2. By a standard density argument we mayassume that u0 ∈ S(M). Here S(M) are the Shwartz functions on M. For the caseM = Td, S(M) = C∞(M). Given a n ∈ N we want to construct a solution un andtn ∈ (0, 1

n) such that

‖un(0)− u0‖Hs ≤ 1

nbut ‖un(tn)‖Hs > n. (1.14)

Main Idea: Since s < 0 contributions to low frequencies are weighted heavily in calculatingthe Hs norm of a function. With this in mind we aim to exploit the following interaction

(High ∼ N)× (High ∼ 2N)× (High ∼ N)→ Low.

for N = N(n) 1. Recall that

Fx(u1u2u3)(ξ) =∑

ξ=ξ1−ξ2+ξ3

u1(ξ1)u2(ξ2)u3(ξ3). (1.15)

Given n ∈ N, let N = N(n) 1. Set

φn(ξ) = R (χNe1+QA(ξ) + χNe1+QA(ξ))

where QA = [−A2 ,

A2 ), A N . Choose R and A large enough so that

RAd ‖u0‖FL1 .

Note that‖φn‖FL1 ∼ RAd, ‖φn‖Hs ∼ RAd/2N s.

Set u0,n = u0+φn. We will study the corresponding solutions un of (1.1) with un|t=0 = u0,n.

4 W. J. TRENBERTH

1.4. Power Series Expansion of un. Recall that we have the expansion

un =

∞∑

j=0

Ξj(u0,n). (1.16)

Lemma 1.4 guarantees absolute and uniform convergence on [−T, T ] if

T .(‖u0‖FL1 +RAd

)−2∼ (RAd)−2. (1.17)

Note also that

‖u0,n − u0‖Hs = ‖φn‖Hs ∼ RA d2N s (1.18)

‖Ξ0(u0,n‖Hs ≤ ‖φn‖Hs + ‖φn‖Hs ∼u0 1 +RAd/2N s. (1.19)

(1.20)

Lemma 1.5 and 1.17 imply

‖Ξ1(u0,n)(t)− Ξ1(φn)(t)‖Hs . t‖u0‖L2

(‖u0‖FL1 + ‖φn‖2FL1

). t‖u0‖L2R2A2d. (1.21)

Lemma 1.6. Let s < 0. Then,

‖Ξj(u0,n)(t)‖Hs ≤ Cjtj(RAd)2j (Rf(A) + ‖u0‖L2)

where here

f(A) =

1, s < −d2

(logA)12 , s = −d

2

Ad2

+s, s > −d2 .

Proof. Recall that supp φn is the disjoint union of two cubs of volume Ad. Also note thatfor T ∈ T (J), Ψφn(T ) is the 2j + 1-fold product of S(t)φn and its c.c. This implies that

suppFx(Ψφn(T )) is contained in at most 22j+1 cubes of volume Ad. Since s < 0, 〈ξ〉s isdecreasing and so

‖〈ξ〉s‖L2(suppFx(Ψφn (T )) ≤ ‖〈ξ〉s‖L2ξ(C

jQA) .

1, s < −d2

Cj(logA)12 , s = −d

2

CjAd2

+s, s > −d2 .

Hence using Holders inequality and then Lemma’s 1.4 and 1.5 we have

‖Ξj(φn)(t)‖Hs ≤∑

T ∈T (j)

‖Ψφn(T )(t)‖Hs

≤∑

T ∈T (j)

‖〈ξ〉s‖L2(suppFx(Ψφn (T ))‖Ψφn(T )(t)‖FL∞

≤ Cjtj(RAd)2j − 1(RAd2 )2f(A).

Similarly one can prove

‖Ξj(u0 + φn)(t)− Ξj(ϕn)(t)‖Hs ≤ Cjtj‖u0‖L2

(‖u0‖2jFL1 + ‖φn‖2jFL1

). (1.22)

Putting this together one gets the result. Proposition 1.7. Let 0 < t N−2. Then,

‖I[S(t)φn, S(t)φn, S(t)φn]‖ = ‖Ξ1(φn)(t)‖Hs & tR3A2df(A).


Proof. Recall the definition of Ξ1(φn)(t) = I[S(t)φn, S(t)φn, S(t)φn],

Ξ1(φn)(t) =

ˆ t

0S(t− t′)

(S(t′)φS(t′)φ§(t′)φ

)dt′.

On the Fourier side,

Fx (Ξ1(φn)(t)) (ξ) = e−it|ξ|2

ˆ

ξ=ξ1−ξ2+ξ3

ˆ t

0eit′µ(ξ) dt′

φn(ξ1)φn(ξ2)φn(ξ3)dξ1dξ2dξ3.

Where µ(ξ) = |ξ|2 − |ξ1|2 + |ξ2|2 − |ξ3|2. As ξi . N for i = 1, 2, 3, it follows that |ξ| . Nand so

t′µ(ξ) << 1 for 0 < t N−2 =⇒ Re

ˆ t

0eit′µ(ξ) dt′.

Using the fact thatχa+QA ∗ χb+QA(ξ) ≥ cdAdχa+b+QA

we then get|Fx (Ξ1(φn)(t)) (ξ)| & tR3A2dχQA(ξ).

The result follows with ‖〈ξ〉s‖L2ξ(QA) ∼ f(A) as before.

1.5. Putting Things Together. We now go back to the proof of Theorem 1.2.

Proposition 1.8. Suppose that for N(n) q,

(1) RAd2N s 1

n

(2) TR2A2d 1(3) TR3A2df(A) n(4) TR3A2df(A) T 2R5A4df(A)(5) T N−2

(6) RAd ‖u0‖FL1, Rf(A) ‖u0‖L2.

Then

‖u0,n − u0‖Hs <1

nbut ‖un(tn)‖Hs > n.

Proof. As ‖u0,n‖FL1 ∼ RAd from condition 2 un exists on [−T, T ] and the power seriesexpansion converges in CTFL1

x and by Lemma 1.6 and summing a geometric series we have

‖∞∑

j=2

Ξj(u0,n)(T )‖Hs . T 2R4A2d (Rf(A) + ‖u0‖L2) ∼ T 2R5A2df(A).

Further

‖un(T )‖Hs ≥ ‖Ξ1(φn)(T )‖Hs − ‖Ξ0(u0,n)(T )‖Hs − ‖Ξ0(u0,n)(T )− Ξ0(φn)(T )‖

− ‖∞∑

j=2

Ξj(u0,n)(T )‖Hs .

Using (1.19), (1.21) and (1.5) to get a lower bound on the negative terms in the aboveinequality we have,

‖un(T )‖Hs & TR3A2df(A)− (1 +RAd2N s)− TR2A2d‖u0‖L2 − T 2R5A4df(A)

∼ TR3A2df(A) n

6 W. J. TRENBERTH

where the last inequality is due to condition 3. This completes the proof It now remains to verify conditions 1 through 6. We do this in the three separate cases

of s < −d2 , s = −d

2 and s > −d2 .

Case 1 s < −d2 : In this case f(A) = 1. We set

A = N1d

(1−σ), R = N2σ, T = N−2−3σ where σ 1.

ThenRA

d2N s = N s+ 1

2+ 3

2σ n

for N = N(n) large enough as the power of N is negative. This verifies condition 1. Forcondition 2 we calculate

TR2A2d = N−σ 1

for N large enough. For condition 3 we have,

TR3A2d = Nσ n

choosing N = N(n) large enough. Conditions 4, 5 and 6 follow in a more immediate fashion

Case 2 s = −d2 : In this case f(A) ∼ (logA)

12 . We set

A =N

1d

(logN)1

16d

, R = 1, T =1

N2(logN)18

Then,

RAd2N s = N

12

(1−d)(logN)−132 1

nfor N = N(n) large enough noting that the power of N is negative. This verifies condition1. For condition 2 we have

TR2A2d = (logN)−14 1

for N large enough. For condition 3 we have,

TR3A2d(logA)12 ∼ (logN)−

14

(logN − 1

16log logN

) 12

∼ (logN)14 n

for N = N(n) large enough.

Case 3 s > −d2 : This case is relevant only for d ≥ 2. In this case f(A) ∼ A d

2+s. We set

A = N2d−σ, R = N−1−s+ d

2σ−θ, T = N−2+2s+dσ+θ

where σ θ > 0 are small such that −2s > dσ + θ and −sσ > 2θ. As before we have

RAd2N s = N−θ 1

n

for N = N(n) large enough which verifies condition 1. For condition 2 we have

TR2A2d = N−θ 1.

Finally for condition 3 we have

TR3A2d ·A d2

+s = N(−d+2d )s−2θ−sσ ≥ N−2θ−sσ n

where we used the fact d ≥ 2 =⇒(−d+2

d

)s is positive and that −2θ − sσ is positive.

Conditions 4, 5 and 6 follows similarly.

This completes the proof of Theorem 1.2.


Remark 1.9. In [1] Ill-posedness for the NLS on Rd was proven for s < −d2 or 0 < s < scrit.

The range −d2 ≥ s < 0 was left open. They used an ODE argument using that for a fixed

the solution toi∂tu+ |u|u2u = 0

isu(t, x) = eit|u0(x)|2u0(x)

and so‖u(t)‖Hs ∼u0 ts, s > 0.

This was then combined with scaling to give the result.

References

[1] M. Christ, J. Colliander, T. Tao, Ill-posedness for nonlinear Schrodinger and wave equations,arXiv:math/0311048 [math.AP].

[2] Oh, T. A remark on norm inflation with general initial data for the cubic nonlinear Schrdinger equationsin negative Sobolev spaces Funkcialaj ekvacioj-Serio internacia, 60(2), 259-277. DOI: 10.1619/fesi.60.259

NONUNIQUENESS OF NLS BELOW L2(T)

TYPED BY JUSTIN FORLANO

Abstract. These notes are based on part of lecture 19 and all of lecture 20.

1. Introduction

In the previous lectures, we have studied the well-posedness theory of the cubic nonlinearSchrodinger equation (NLS):

i∂tu+ ∂2xu± |u|2u = 0

u|t=0 = u0,(t, x) ∈ R+ ×M, (1.1)

withM = R or T. By the algebra property of Hs(M), we proved that NLS (1.1) is locallywell-posed (LWP) when s > 1

2 . Exploiting the dispersive properties of (1.1), we saw that

on R the Strichartz estimates implied LWP down to L2(R). More precisely, we showedthat for any u0 ∈ L2(R), there exists a time T = T (u0) > 0, such that we have a solutionu ∈ CTL2(R) ∩ L8

TL4x(R) ⊂ CTL2(R) which is unique within the restricted class

CTL2(R) ∩ L8

TL4x(R),

but not necessarily on the whole of CTL2(R). On T, Bourgain [1] introduced the Fourier

restriction spaces Xs,b adapted to the Schrodinger equation and used them to prove LWPin Hs(T) for any s ≥ 0. The solutions constructed are unique only within the restricted

class Xs, 12+ ⊂ CTHs.

We will focus now on the case when M = T.From lecture 16 onwards, we studied ill-posedness results for NLS (1.1). For instance,

Christ-Colliander-Tao [3] showed the solution map in Hs(T), s < 0, is discontinuous atthe origin, and Guo-Oh [4] proved non-existence of solutions from initial data belonging toHs(T) \ L2(T) with s ∈

(− 1

8 , 0). It was argued in the previous lectures that for studying

negative Sobolev regularity problems on T, a more appropriate model than NLS (1.1) isthe Wick ordered (renormalized) cubic nonlinear Schrodinger equation (WNLS):

i∂tu+ ∂2xu+N (u) = 0

u|t=0 = u0,(t, x) ∈ R+ × T, (1.2)

where N is the Wick ordered cubic nonlinearity defined as

N (u) = |u|2u− 2

(

T|u|2dx

)u

“ = |u|2u− 2 · ∞ · u”, if u /∈ L2x(T).

(1.3)

Note that

N (u)(n) =∑

n=n1−n2+n3n1,n3 6=n

un1 un2 un3 − |un|2un. (1.4)

1

2 J. FORLANO

When n1 = n or n3 = n,

∑

n=n1−n2+n3

un1 un2 un3 =

(

T|u|2dx

)un,

which is infinite if u /∈ L2x(T). It is precisely these two contributions which are removed to

create (1.4) and this explains the factor of ‘2’ in (1.3).In light of the existence theory described above, it is natural to ask whether uniqueness

holds in the whole space CTHs, without intersecting with some auxiliary spaces. Such

a notion of uniqueness is known as unconditional uniqueness. Consequently, solutions areindependent of how they are constructed and so we can genuinely compare solutions derivedfrom different methods.

An immediate problem arises though if one considers general solutions belonging to

CTHs. In order to make sense of N (u) as a space-time distribution for general u, we need

at least u(t) ∈ L3loc(T). Sobolev embeddings imply it would suffice to have s ≥ 1

6 . Although

the solutions constructed by Bourgain [1] were of regularity < 16 , defining the nonlinearity

was not an issue as the periodic L4-Strichartz estimate he proved implied the solutionsalso belonged to L3

x(T). Therefore, in order to even pose the question of unconditionaluniqueness for (1.2) in low regularity, we must first make sense of the nonlinearity.

Christ [2] gave a meaningful notion of the existence of the nonlinearity for rough functionsand proved nonuniqueness of solutions to the renormalized NLS (1.2) in CTH

s, for anys < 0, within the sense he introduced. We describe this notion of solution below andrefer to Theorem 1.4 for a precise statement of his result. His result essentially says thatwe cannot expect unconditional uniqueness to hold for negative regularities. It is worthpointing out that the solution constructed in Theorem 1.4 does not belong to the Fourierrestriction spaces Xs,b. Miyaji-Tsutsumi [5] showed that unconditional uniqueness for therenormalized third order dispersion cubic nonlinear Schrodinger equation also fails.

We will now describe Christ’s notion of the nonlinearity.

Definition 1.1. Let PNN∈N denote a family of Fourier cutoff operators defined by

PNf(n) = mN (n)f(n),

where mNN∈N is a uniformly bounded family of functions mN : Z→ C of finite supportthat satisfies

limN→∞

mN (n) = 1, for all n ∈ Z.

Definition 1.2. Let u ∈ D′((0, T )×T). We say N (u) exists and is equal to v ∈ D′((0, T )×T) if

limN→∞

N (PNu) = v in D′((0, T )× T), (1.5)

for any sequence PNN∈N of Fourier cutoff operators.

Notice that for each fixed N , N (PNu) is always well-defined as a distribution becausePNu ∈ C∞x . Notice also that we require (1.5) to hold for every Fourier cutoff operator.For rough u, we define the nonlinearity N (u) in (1.2) as the limit of the nonlinearity forsmoothed u as in Definition 1.2.

Definition 1.3. We say u ∈ CTHs is a weak solution to (1.2) in the extended sense withu|t=0 = u0 if N (u) exists in the sense of Definition 1.2 and u satisfies (1.2) in the sense ofdistributions.

NONUNIQUENESS OF NLS 3

Recall that F ∈ C−1T Hs if´ t0 F (t′)dt′ ∈ CTHs and

‖F‖C−1T Hs := max

t∈[0,T ]

∥∥∥∥ˆ t

0F (t′)dt′

∥∥∥∥Hs

. (1.6)

Theorem 1.4 (Christ [2]). Let s < 0. There exists u ∈ CTHs not identically zero such thatu is a weak solution to (1.2) in the extended sense with u|t=0 ≡ 0. Moreover, S(−t)N (u) ∈C−1T Hs.

Remark 1.5. The proof of Theorem 1.4 can be adapted to give a similar nonuniquenessresult for (1.2) in the Fourier-Lebesgue spaces FLp(T), for p > 2.

Remark 1.6. The proof also applies to quadratic nonlinearities Q(u) of the form

Q(u) = u2, u2, |u|2 −(

T|u|2dx

).

The rest of these notes are devoted to the proof of Theorem 1.4 as given in [2].

2. Idea of the proof

The heuristic idea of the proof of Theorem 1.4 is to drive the high to low frequencycascade by the introduction of a forcing F in the inhomogeneous problem

i∂tu+ ∂2xu+N (u) = F, on T. (2.1)

If F is supported on high-frequencies, it forces high frequency responses in the solution u.The cubic nonlinearity N (u) then transfers high frequency interactions to low frequencies(via the h × h × h → l cascade from the previous lectures) and as s < 0, the Hs-normof the nonlinearity will become large. As the forcing F is driven to higher and higherfrequencies, the effect of the nonlinearity increases because of the high-to-low cascade whilesimultaneously F itself tends weakly to zero by virtue of its fast oscillations. Then in thelimit of large driving frequency, we hope to construct a non-zero solution to (1.2).

To implement this procedure, we consider a sequence of smooth solutions u(n)n∈N tothe inhomogeneous equation

i∂tu(n) + ∂2xu

(n) +N (u(n)) = F (n),

with a sequence of driving forces F (n) supported on frequencies |k| ≥M (n), where M (n) →∞ as n→∞. We write u(n) = v(n) + f (n), where

f (n)(t) := −iˆ t

0S(t− t′)F (n)(t′)dt′,

and consider the equation satisfied for vn:

i∂tv(n) + ∂2xv

(n) +N (v(n)) + (N (v(n) + f (n))−N (v(n))−N (f (n))) = −N (f (n)). (2.2)

The construction proceeds by choosing F (n) such that N (f (n)) ∼ F (n−1) so that f (n) canbe made small in CtH

s. This should then imply the difference term on the left hand side of(2.2) can be considered a small error. At the same time, N (f (n)) is made large in C−1t Hs

so that vn is essentially driven by F (n) and the high-to-low frequency cascade is initiated.The goal then is to show that in the limit n → ∞, f (n) vanishes and v(n) converges to anon-zero weak solution (in the extended sense) of the homogeneous nonlinear equation.

Remark 2.1. The works of Scheffer [7] and Shnirelman [8] also obtained nonuniquenessresults for the incompressible Euler equation through studying the inhomogeneous problem.

4 J. FORLANO

The key result in the construction is the following proposition.

Proposition 2.2. Let s < 0 and suppose u ∈ C∞([0, T ] × T) such that for all n ∈ Z,u(t, n)→ 0 at infinite order as t→ 0+. Then, for all ε > 0, there exist vε, Fε ∈ C∞([0, T ]×T) whose Fourier coefficients decay to zero at infinite order as t → 0+ such that vε is asolution to (2.1) with bounds

‖vε − u‖CTHs ≤ ε,‖S(−t)Fε‖C−1

T Hs ≤ ε.

We split the proof of Theorem 1.4 into three parts:

(1) The proof of Proposition 2.2,(2) construction of the non-zero candidate solution,(3) verifying this candidate indeed solves (NLS).

3. First step: Proof of Proposition 2.2

In terms of the Fourier coefficients u(t, n)n∈Z, (2.1) can be expressed as the followinginfinite dimensional system of ODEs:

i∂tun − in2un +∑

n=n1−n2+n3n6=n1,n3

un1 un2 un3 − |un|2un = Fn(t). (3.1)

In terms of the interaction representation

yn(t) = FxS(−t)u(t)(n) = eitn2u(t, n),

(3.1) becomes

∂tyn = i∑

n=n1−n2+n3n6=n1,n3

eiφ(n)tyn1yn2yn3

︸︷︷︸=:NR(y)=non-resonant

−i|yn|2yn︸︷︷︸=:R(y)=resonant

−ie−itn2Fn(t). (3.2)

Here

φ(n) = φ(n1, n2, n3, n) = n2 − n21 + n22 − n23= 2(n− n1)(n− n3),

where the third equality above holds when n = n1−n2+n3. With a slight abuse of notation,we will sometimes consider NR as a trilinear operator, that is,

NR(v, w, z) = i∑

n=n1−n2+n3n6=n1,n3

eiφ(n)tvn1wn2zn3 ,

with NR(y, y, y) =: NR(y). Considering the sequence yn(t)n∈Z as an infinite-dimensional vector y(t), (3.2) is further written as

dy

dt= NR(y) +R(y)︸︷︷︸

=:N (y)=full nonlinearity

+f, (3.3)

where f has components fn := −ie−itn2Fn(t).

Definition 3.1. We say the sequence-valued function x(t) = xn(t)n∈Z has support inS ⊂ Z if xn(t) ≡ 0 for all n 6∈ S and all t ∈ [0, T ].


Define ‖a‖`2s := ‖F−1a‖Hs =(∑

n∈Z〈n〉2s|an|2) 1

2 .In this section, we prove the following result which is just a rewrite of Proposition 2.2

but on the frequency side (and hence implies the aforementioned proposition).

Proposition 3.2. Let s < 0. Let x ∈ C∞([0, T ]) be a finitely-supported sequence-valuedfunction which vanishes at infinite order as t → 0+. Then, for all ε > 0, there existssequence-valued functions y and g ∈ C∞([0, T ]) with finite support such that

∂ty = N (y) + g,

y(t)→ 0 at infinite order as t→ 0+ and we have

‖y − x‖CT `2s≤ ε, (3.4)

‖g‖C−1T `2s≤ ε. (3.5)

Moreover, for any M > 0, y − x and g can be constructed to have support in [M,∞).

Proof. Let

f = ∂tx−N (x). (3.6)

As x has finite support, so does N (x) and hence f has finite support which we denoteby S := nj : 1 ≤ j ≤ A, where we have enumerated its elements. We construct a

finite set S ⊂ Z ∩ [M,∞) as follows: Pick m1 ≥ M and set m′1 by 2m1 −m′1 = n1. Bychoosing1 m1 1, we ensure that m′1 ≥ M . Next pick m2 m1,m

′1 and set m′2 such

that 2m2 −m′2 = n2. By choosing m2 1, we ensure m2 ≥ 1 and also m′2 m1,m′1.

Repeating this process for each 1 ≤ j ≤ A, gives us a finite set S = m1,m′1, . . . ,mA,m

′A

whose elements satisfy

2mj −m′j = nj , for all 1 ≤ j ≤ A (=⇒ m′j ≈ 2mj), (3.7)

subject to the following additional constraints:

(1) For any k, `,m ∈ S such that ` 6= k,m, we have |k − `+m| ≥M unless (k, `,m) =(mj ,m

′j ,mj) for some j,

(2) For any k, ` ∈ S and n ∈ suppx, we have |k − n + `| ≥ M and |k − ` + n| ≥ Munless k = `,

(3) For any k ∈ S and m,n ∈ suppx, |k −m+ n|, |m+ k − n| ≥M .

The first constraint can be satisfied because a given mk (and m′k) is much larger than all

previous mj (j < k). Recall as well that all the elements in S are positive. The other

conditions are also reasonably satisfied by construction of S.Construct C∞ functions hm(t) : m ∈ S that vanish at infinite order as t → 0+

satisfying

ihmj (t)hm′j (t)hmj (t) =1

2e−iφ(mj ,m

′j ,mj ,nj)tfnj (t), (3.8)

and set hm(t) ≡ 0 for all m /∈ S. When m ∈ S, the functions hm(t) will in gen-eral not be unique, but from (3.8), they can be chosen so that maxm∈S ‖hm‖C([0,1]) .maxn∈S ‖fn‖

13

C([0,1]) (very roughly!) which is a bound only depending on S and f and not

on S. Notice from (3.8) that any C1([0, 1]) bound of hm will depend on S. A consequenceof this is we cannot make dhm

dt small in CT `2s.

1In particular, we take m1 n1 so that m′1 ≈ 2m1. This is to mimic the h× h× h→ l interaction, e.g.m1 − 2m1 + m1 = 0.

6 J. FORLANO

Let

y = x+ h. (3.9)

The interpretation is that x carries the low frequencies while h carries the high frequencies,and since s < 0, the contribution from h will be small. Indeed, we have

‖hm‖CT `2s.M s|#S| 12 max

m∈S‖hm‖CT

≤ CM s, (3.10)

where C is a constant only depending on f and S and hence only on x. Choosing Msufficiently large, (3.9) and (3.10) imply (3.4).

Let

g =dy

dt−N (y).

As x and h have disjoint supports,

R(y) = R(x+ h) = R(x) +R(h). (3.11)

Thus by the definition of N from (3.3), (3.6), (3.9) and (3.11), we have

g = (f −NR(h)) +dh

dt−R(h)− (NR(x+ h)−NR(x)−NR(h)). (3.12)

Wishing to show (3.5), we will separately bound the C−1T `2s-norms of each quantity on theright hand side of (3.12).

• ‖f −NR(h)‖C−1T `2s

: It suffices to bound ‖f −NR(h)‖CT `2s, in which case we study,

for fixed t ∈ [0, T ],∑

n

〈n〉2s|fn −NR(h)(n)|2 =∑

n∈S〈n〉2s|fn −NR(h)(n)|2 +

∑

n/∈S〈n〉2s|NR(h)(n)|2.

By definition of the functions hm, the triples (m1,m2,m3) we sum over inNR(h)(n)

are restricted to (S)3. Then by (3.7) and (3.8), for each fixed n ∈ S, f(n) −NR(h)(n) ≡ 0, so the first term is identically zero. For the second term, condition(i) implies NR(h) is supported on a finite number of frequencies in the set n :

|n| ≥M, depending on |#S|. Therefore, we obtain

‖f −NR(h)‖CT `2s≤ CM s. (3.13)

•∥∥dhdt

∥∥C−1

T `2s: By (1.6) and (3.10), we have

∥∥∥∥dh

dt

∥∥∥∥C−1

T `2s

≤ CM s. (3.14)

• ‖R(h)‖C−1T `2s

: As h is supported on [M,∞), then

‖R(h)‖C−1T `2s≤ ‖R(h)‖CT `2s

≤ ‖h‖2CT `∞‖h‖CT `2s≤ CM s. (3.15)

• ‖NR(x+h)−NR(x)−NR(h)‖C−1T `2s

: Again, it suffices to bound in CT `2s. For each

fixed n, expanding NR(x+h)(n)−NR(x)(n)−NR(h)(n) yields sums of the formNR(x, x, h) and NR(x, h, h). It is then clear from conditions (2) and (3) that theset of allowable frequencies with non-vanishing contribution to the `2s summation isfinite and supported on |n| ≥M . Thus

‖NR(x+ h)−NR(x)−NR(h)‖CT `2s≤ CM s. (3.16)


Combining (3.12), (3.13), (3.14), (3.15) and (3.16), we have

‖g‖C−1T `2s≤ CM s ≤ ε,

for M sufficiently large since s < 0. This verifies (3.5) which completes the proof.

4. Second step: Construction of the non-zero candidate

In this step, we construct a sequence of C∞ sequence-valued functions x(n), which vanishto infinite order as t → 0+, and converge to a limit in CT `

2s. This limit, denoted by x,

will be non-zero in CT `2s and thus u = F−1e−itn2

x will be our candidate solution for theclaim of Theorem 1.4.

Let x(0) be a C∞ sequence-valued function of finite support, vanishing at infinite orderas t→ 0+, and satisfying

‖x(1)0 ‖CT≥ 1, (4.1)

that is, the zeroth frequency mode of x(1) is not the zero element in CT . To construct thesequence x(n), we proceed inductively in n ∈ N. Viewing x = x(n), we apply Proposition3.2 to construct x(n+1) = y. Setting h(n) = y − x = x(n+1) − x(n) and f (n+1) = ∂tx

(n+1) −N (x(n+1)), we have that h(n) vanishes to infinite order as t → 0+, and hence so too does

x(n+1). Further, for ε > 0 sufficiently small in our applications of Proposition 3.2, we have

‖h(n)‖CT `2s≤ εn, (4.2)

‖f (n+1)‖C−1T `2s≤ εn, (4.3)

where εn ≤ 2−n−1. It is clear by construction that

x(n) = x(1) +n−1∑

j=1

h(j), (4.4)

and thus, (4.2) implies x(n)n∈N is Cauchy in CT `2s. Set

x = limn→∞

x(n) in CT `2s, (4.5)

and note that (4.4) implies

x = x(1) +

∞∑

j=1

h(j). (4.6)

It follows from (4.6) that x 6≡ 0 since (4.1) and (4.2) imply

‖x0‖CT≥ ‖x(0)0 ‖CT

−∞∑

j=1

‖h(j)‖CT≥ 1

2. (4.7)

Finally, as x(n)(0) ≡ 0 for every n, then x(0) ≡ 0.

8 J. FORLANO

5. Third step: Verify candidate is a solution

In this final step, we will show that the candidate x indeed solves (3.3) (with f ≡ 0) and

that this implies u = F−1e−itn2x is a non-zero weak solution in the extended sense to

(1.2), which completes the proof of Theorem 1.4.Let mNN be a sequence of uniformly bounded, finitely-supported functions from Z to

C such that, for any n ∈ Z, limN→∞mN (n) = 1. With a slight abuse of notation, we setPNan = mN (n)an. Note that the operators PN are uniformly bounded in N on the spacesCTX and C−1T X, where X = `2s or `1.

As stipulated by Definitions 1.2 and 1.3, it remains to prove (in this order):

(I) limN→∞N (PNx) exists in C−1T `2s for any sequence PNN of Fourier cut-off opera-tors as defined above. This implies N (x) exists in the weak sense.

(II) N (x(n))→ N (x) as n→∞ in C−1T `2s.

(III) u = F−1e−itn2x is a solution to (1.2) in the extended sense.

Suppose we have verified (I) and (II) and wish to show (III). By construction, for each

n ∈ N, x(n) solves (3.6) so that

x(n)(t) = 0 +

ˆ t

0N (x(n)(t′))dt′ +

ˆ t

0f (n)(t′)dt′. (5.1)

By (4.3), f (n) → 0 in C−1T `2s. By (4.5), x(n) → x in C−1T `2s and by (I) and (II), N (x(n))→N (x) in C−1T `2s. Therefore (5.1) implies

x(t) = 0 +

ˆ t

0N (x)(t′)dt′. (5.2)

Now (I) equivalently shows that N (u) exists while (4.7) shows u(t) 6≡ 0. Clearly u(0) ≡ 0.Thus u is a weak solution to (1.2) in the extended sense such that u|t=0 = 0 but u(t) 6≡ 0for t ∈ (0, T ], where T > 0.

Let us now verify (I).By Young’s inequality, we have

‖N (v)−N (w)‖CT `1 . (‖v‖2CT `1+ ‖w‖2CT `1

)‖v − w‖CT `1 . (5.3)

For any N and n, (5.3) implies

‖N (PNx(n))−N (PN (x))‖CT `1 . ‖PN (x(n) − x)‖CT `1(‖PNx(n)‖2CT `1

+ ‖PNx‖2CT `1)

. N3‖x(n) − x‖CT `∞

. N32−n, (5.4)

where we have used that PNw is supported on frequencies . N for any w, and that (4.2)and (4.6) imply

‖x(n) − x‖CT `∞ . 2−n.It follows from (5.4) that for any J and fixed N , we have

N (PNx) = N (PNx(J)) +

∞∑

j=J

(N (PNx(j+1))−N (PNx

(j))) (5.5)

in CT `1. As limN→∞mN (n) = 1, then for any fixed n, N (PNx

(n)) → N (x(n)) in CT `1 as

N →∞. Therefore (5.3) implies

limN→∞

‖N (PNx(n))−N (x(n))‖CT `1 = 0. (5.6)


We require one last result in order to conclude (I).

Lemma 5.1. For any n and N , there exists a constant C < ∞ independent of n and Nsuch that

‖N (PNx(n+1))−N (PNx

(n))‖C−1T `2s≤ C2−n. (5.7)

Proof. Recalling that x(n+1) = x(n) + h(n), we write

N (PNx(n+1))−N (PNx

(n)) =(N (PNx(n) + PNh

(n))−N (PNx(n))−N (PNh

(n)))

+NR(PNh(n)) +R(PNh

(n)),

and estimate each of these separately in C−1T `2s.

• ‖R(PNh(n))‖C−1

T `2s: From (3.2), (3.10) and (4.2), we have

‖R(PNh(n))‖C−1

T `2s. ‖mN‖2`∞‖PNh(n)‖C−1

T `2s. ‖mN‖3`∞2−n.

• ‖N (PNx(n) +PNh


(n))‖C−1T `2s

: As the difference contains

terms all with at least one occurrence of h(n), we get from (4.4) and (4.2)

‖N (PNx(n) + PNh


(n))‖C−1T `2s. ‖mN‖3`∞2−n.

• ‖NR(PNh(n))‖C−1

T `2s: As in the proof of (3.13), we split the `2s norm into a sum

over S and its complement. For the sum over frequencies in S, which correspondto triples (mj ,m

′j ,mj) ∈ (S)3 for the sum NR(PNh

(n)), (3.8) furnishes us a bound

by ‖mN‖3`∞‖f‖C−1T `2s

at which point, (4.3) implies the desired estimate by 2−n. For

the sum over frequencies not in S (triples (k, `,m) ∈ (S)3 with ` 6= k,m and notriples (mj ,m

′j ,mj)), we argue as in the similar term for (3.13).

Now we show that the sequence N (PNx)N∈N is Cauchy in C−1T `2s. Given ε > 0, chooseJ ≥ 1 so large so that

C2−J <ε

3. (5.8)

By (5.6), we can find N0 ≥ 1 such that for any N,M ≥ N0 , we have

‖N (PNx(J))−N (PMx

(J))‖C−1T `2s

<ε

3. (5.9)

Fix N,M ≥ N0. By (5.5) and Minkowski’s integral inequality, we write

‖N (PNx)−N (PMx)‖C−1T `2s≤ ‖N (PNx

(J))−N (PMx(J))‖C−1

T `2s

+

∞∑

j=J

‖N (PNx(j+1))−N (PNx

(j))‖C−1T `2s

+

∞∑

j=J

‖N (PMx(j+1))−N (PMx

(j))‖C−1T `2s

= (1) + (2) + (3).

10 J. FORLANO

By Lemma 5.1 and (5.8), we have

(2) + (3) ≤ 2

∞∑

j=J

C2−j ≤ 2C2−J <2ε

3.

Combining this with (5.9) shows N (PNx)N∈N is Cauchy in C−1T `2s. Hence, N (x) is well-

defined and belongs to C−1T `2s.Finally, we show (II). By (I) and (5.6), we have

N (PNx)−N (PNx(n))→ N (x)−N (x(n)) in C−1T `2s as N →∞.

Now Lemma 5.1 implies

‖N (PNx)−N (PNx(n))‖C−1

T `2s≤ C2−n,

uniformly in N . Thus

‖N (x)−N (x(n))‖C−1T `2s

= limN→∞

‖N (PNx)−N (PNx(n))‖C−1

T `2s≤ C2−n.

Taking n→∞ completes the proof of (II) and hence the proof of Theorem 1.4.

References

[1] J. Bourgain, Fourier transform restriction phenomena for certain lattice subsets and applications tononlinear evolution equations. I. Schrodinger equations, Geom. Funct. Anal. 3 (1993), 107–156.

[2] M. Christ, Nonuniqueness of weak solutions of the nonlinear Schrodinger equation, arXiv:math/0503366[math.AP].

[3] M. Christ, J. Colliander, T. Tao, Ill-posedness for nonlinear Schrodinger and wave equations,arXiv:math/0311048 [math.AP].

[4] Z. Guo, T. Oh Ill-posedness for nonlinear Schrodinger and wave equations, Internat. Math. Res. Not.(2016). doi: 10.1093/imrn/rnw271.

[5] T. Miyaji, Y. Tsutsumi Local well-posedness of the NLS equation with third order dispersion in negativeSobolev spaces, Differential Integral Equations 31 (2018), no. 1-2, 111–132.

[6] T. Oh, A remark on norm inflation with general initial data for the cubic nonlinear Schrodinger equa-tions in negative Sobolev spaces, Funkcial. Ekvac. 60 (2017), 259–277.

[7] V. Scheffer, An inviscid flow with compact support in space-time, J. Geom. Anal. 3 (1993), no. 4,343–401.

[8] A. Schnirelman, On the nonuniqueness of weak solution of the Euler equation, Comm. Pure Appl. Math.50 (1997), no. 12, 1261–1286.

migsaa advanced ph.d. course

Documents